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1.6 inverse function (optional)

The document discusses inverse functions. It begins by defining a function f(x) that takes an input x and produces an output y. It then introduces the concept of finding the inverse of this function - the reverse procedure that takes the output y and finds the corresponding input(s) x. This reverse procedure may or may not be a function. For it to be a function, f(x) must be one-to-one, meaning that different inputs x produce different outputs y. The document provides examples and diagrams to illustrate one-to-one functions and their inverses. It concludes that if f(x) is one-to-one, then its inverse function f-1(y) is well-defined

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Inverse Functions
Inverse Functions A function f(x) = y takes an input x and produces one output y.
Inverse Functions A function f(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x?
Inverse Functions A function f(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x? This procedure of associating the output y to the input x may or may not be a function.
Inverse Functions A function f(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x? This procedure of associating the output y to the input x may or may not be a function. If it is a function, it is called the inverse function of f(x)
Inverse Functions A function f(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x? This procedure of associating the output y to the input x may or may not be a function. If it is a function, it is called the inverse function of f(x) and it is denoted as f -1(y).
Inverse Functions A function f(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x? This procedure of associating the output y to the input x may or may not be a function. If it is a function, it is called the inverse function of f(x) and it is denoted as f -1(y). We say f(x) and f -1(y) are the inverse of each other.
Inverse Functions A function f(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x? This procedure of associating the output y to the input x may or may not be a function. If it is a function, it is called the inverse function of f(x) and it is denoted as f -1(y). We say f(x) and f -1(y) are the inverse of each other. Example A. Let f(x) = x2 = y. Suppose y = 9, what is (are) the input x that produces y = 9? Is this reverse procedure a function?
Inverse Functions A function f(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x? This procedure of associating the output y to the input x may or may not be a function. If it is a function, it is called the inverse function of f(x) and it is denoted as f -1(y). We say f(x) and f -1(y) are the inverse of each other. Example A. Let f(x) = x2 = y. Suppose y = 9, what is (are) the input x that produces y = 9? Is this reverse procedure a function? Since f(x) = x2 = 9,
Inverse Functions A function f(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x? This procedure of associating the output y to the input x may or may not be a function. If it is a function, it is called the inverse function of f(x) and it is denoted as f -1(y). We say f(x) and f -1(y) are the inverse of each other. Example A. Let f(x) = x2 = y. Suppose y = 9, what is (are) the input x that produces y = 9? Is this reverse procedure a function? Since f(x) = x2 = 9, so x = ±√9 x = – 3, x = 3.
Inverse Functions A function f(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x? This procedure of associating the output y to the input x may or may not be a function. If it is a function, it is called the inverse function of f(x) and it is denoted as f -1(y). We say f(x) and f -1(y) are the inverse of each other. Example A. Let f(x) = x2 = y. Suppose y = 9, what is (are) the input x that produces y = 9? Is this reverse procedure a function? Since f(x) = x2 = 9, so x = ±√9 x = – 3, x = 3. This reverse procedure takes y = 9 and associates to it two different answers so it is not a function. What condition is needed for a function to have an inverse?
Inverse Functions The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs
Inverse Functions The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9).
Inverse Functions The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9).
A function is one-to-one if different inputs produce different outputs. Inverse Functions The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9).
Inverse Functions The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v).
A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, Inverse Functions a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9).
Inverse Functions u v a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). any pair u = v A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures,
Inverse Functions u f(u) v a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
Inverse Functions u f(u) v f(v) f(u) = f(v) a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
Inverse Functions u f(u) v f(v) f(u) = f(v) a one-to-one function not a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
Inverse Functions u f(u) v f(v) f(u) = f(v) a one-to-one function u v not a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). there exist u = v A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
Inverse Functions u f(u) v f(v) f(u) = f(v) a one-to-one function u v not a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). there exist u = v A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
Inverse Functions u f(u) v f(v) f(u) = f(v) a one-to-one function u v not a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). there exist u = v such that f(u)=f(v) A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
Example B. a. g(x) = 2x is one-to-one Inverse Functions u f(u) v f(v) f(u) = f(v) a one-to-one function u v not a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). there exist u = v such that f(u)=f(v) A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
Example B. a. g(x) = 2x is one-to-one because if u  v, then 2u  2v. Inverse Functions u f(u) v f(v) f(u) = f(v) a one-to-one function u v not a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). there exist u = v such that f(u)=f(v) A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
Example B. a. g(x) = 2x is one-to-one because if u  v, then 2u  2v. b. f(x) = x2 is not one-to-one because for example 3  –3, but f(3) = f(–3) = 9. Inverse Functions u f(u) v f(v) f(u) = f(v) a one-to-one function u v not a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). there exist u = v such that f(u)=f(v) A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
Fact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. Inverse Functions
Fact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function f(x) u
Fact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function u f(u) v f(v) u = v f(u) = f(v) f –1(y) is a well defined function f –1(y)f(x) u
Fact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions Given a simple y = f(x) we may solve equation y = f(x) for x in terms of y to find f –1(y) explicitly. f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function u f(u) v f(v) u = v f(u) = f(v) f –1(y) is a well defined function f –1(y)f(x) u
Fact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions Given a simple y = f(x) we may solve equation y = f(x) for x in terms of y to find f –1(y) explicitly. Example C. Find the inverse function of y = f(x) = x – 5 3 4 f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function u f(u) v f(v) u = v f(u) = f(v) f –1(y) is a well defined function f –1(y)f(x) u
Fact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions Given a simple y = f(x) we may solve equation y = f(x) for x in terms of y to find f –1(y) explicitly. Example C. Find the inverse function of y = f(x) = x – 5 Given y = x – 5, clear the denominator to solve for x.3 4 3 4 f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function u f(u) v f(v) u = v f(u) = f(v) f –1(y) is a well defined function f –1(y)f(x) u
Fact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions Given a simple y = f(x) we may solve equation y = f(x) for x in terms of y to find f –1(y) explicitly. Example C. Find the inverse function of y = f(x) = x – 5 Given y = x – 5, clear the denominator to solve for x. 4y = 3x – 20 3 4 3 4 f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function u f(u) v f(v) u = v f(u) = f(v) f –1(y) is a well defined function f –1(y)f(x) u
Fact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions Given a simple y = f(x) we may solve equation y = f(x) for x in terms of y to find f –1(y) explicitly. Example C. Find the inverse function of y = f(x) = x – 5 Given y = x – 5, clear the denominator to solve for x. 4y = 3x – 20 4y + 20 = 3x 3 4 3 4 f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function u f(u) v f(v) u = v f(u) = f(v) f –1(y) is a well defined function f –1(y)f(x) u
Fact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions Given a simple y = f(x) we may solve equation y = f(x) for x in terms of y to find f –1(y) explicitly. Example C. Find the inverse function of y = f(x) = x – 5 Given y = x – 5, clear the denominator to solve for x. 4y = 3x – 20 4y + 20 = 3x = x 3 4 4y + 20 3 3 4 f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function u f(u) v f(v) u = v f(u) = f(v) f –1(y) is a well defined function f –1(y)f(x) u
Fact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions Given a simple y = f(x) we may solve equation y = f(x) for x in terms of y to find f –1(y) explicitly. Example C. Find the inverse function of y = f(x) = x – 5 Given y = x – 5, clear the denominator to solve for x. 4y = 3x – 20 4y + 20 = 3x = x 3 4 4y + 20 3 3 4 f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function u f(u) v f(v) u = v f(u) = f(v) f –1(y) is a well defined function f –1(y)f(x) u (Note: In general it’s impossible to solve for x explicitly.)
Inverse Functions Let f and f –1 be a pair of inverse functions and that f(a) = b
Inverse Functions Let f and f –1 be a pair of inverse functions and that f(a) = b ba f(a) = b
Inverse Functions Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). ba f(a) = b a = f –1(b)
Inverse Functions Theorem: If f(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b).
Inverse Functions Theorem: If f(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). f(x)x f(x)
Inverse Functions Theorem: If f(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). f(x) f –1(f(x)) = x x f(x)
Inverse Functions Theorem: If f(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). f(x) f –1(f(x)) = x x f(x)
Inverse Functions Theorem: If f(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). f(x) f –1(f(x)) = x x f(x) y f –1(y) f –1(y)
Inverse Functions Theorem: If f(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
Inverse Functions Theorem: If f(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). Example D. Given the pair of inverse functions f(x) = x – 5 and show that f(f –1(y)) = y. 3 4 f –1(y) = 4y + 20 3 f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
Inverse Functions Theorem: If f(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). Example D. Given the pair of inverse functions f(x) = x – 5 and show that f(f –1(y)) = y. f (f –1(y)) = 3 4 f –1(y) = 4y + 20 3 f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
Inverse Functions Theorem: If f(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). Example D. Given the pair of inverse functions f(x) = x – 5 and show that f(f –1(y)) = y. f (f –1(y)) = f ( ) 3 4 f –1(y) = 4y + 20 3 4y + 20 3 f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
Inverse Functions Theorem: If f(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). Example D. Given the pair of inverse functions f(x) = x – 5 and show that f(f –1(y)) = y. f (f –1(y)) = f ( ) = ( ) – 5 3 4 f –1(y) = 4y + 20 3 4y + 20 3 4y + 20 3 3 4 f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
Inverse Functions Theorem: If f(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). Example D. Given the pair of inverse functions f(x) = x – 5 and show that f(f –1(y)) = y. f (f –1(y)) = f ( ) = ( ) – 5 3 4 f –1(y) = 4y + 20 3 4y + 20 3 4y + 20 3 3 4 f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
Inverse Functions Theorem: If f(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). Example D. Given the pair of inverse functions f(x) = x – 5 and show that f(f –1(y)) = y. f (f –1(y)) = f ( ) = ( ) – 5 = – 5 3 4 f –1(y) = 4y + 20 3 4y + 20 3 4y + 20 3 4y + 20 4 3 4 f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
Inverse Functions Theorem: If f(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). Example D. Given the pair of inverse functions f(x) = x – 5 and show that f(f –1(y)) = y. f (f –1(y)) = f ( ) = ( ) – 5 = – 5 3 4 f –1(y) = 4y + 20 3 4y + 20 3 4y + 20 3 4y + 20 4 4(y + 5) 4 = – 5 3 4 f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
Inverse Functions Theorem: If f(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). Example D. Given the pair of inverse functions f(x) = x – 5 and show that f(f –1(y)) = y. f (f –1(y)) = f ( ) = ( ) – 5 = – 5 3 4 f –1(y) = 4y + 20 3 4y + 20 3 4y + 20 3 4y + 20 4 4(y + 5) 4 = – 5 = y + 5 – 5 = y 3 4 f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
Inverse Functions Since we usually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3
Inverse Functions Since we usually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = .
Inverse Functions Since we usually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = . 2x – 1 x + 1 Set f(x) = y = ,
Inverse Functions Since we usually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = . 2x – 1 x + 1 Set f(x) = y = , clear the denominator then solve for x.
Inverse Functions Since we usually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = . 2x – 1 x + 1 y(x + 1) = 2x – 1 Set f(x) = y = , clear the denominator then solve for x.
Inverse Functions Since we usually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = . 2x – 1 x + 1 y(x + 1) = 2x – 1 yx + y = 2x – 1 Set f(x) = y = , clear the denominator then solve for x.
Inverse Functions Since we usually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = . 2x – 1 x + 1 y(x + 1) = 2x – 1 yx + y = 2x – 1 y + 1 = 2x – yx Set f(x) = y = , clear the denominator then solve for x.
Inverse Functions Since we usually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = . 2x – 1 x + 1 y(x + 1) = 2x – 1 yx + y = 2x – 1 y + 1 = 2x – yx y + 1 = (2 – y)x Set f(x) = y = , clear the denominator then solve for x.
Inverse Functions Since we usually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = . 2x – 1 x + 1 y(x + 1) = 2x – 1 yx + y = 2x – 1 y + 1 = 2x – yx y + 1 = (2 – y)x = xy + 1 2 – y Set f(x) = y = , clear the denominator then solve for x.
Inverse Functions Since we usually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 Hence f–1(x) = 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = . 2x – 1 x + 1 y(x + 1) = 2x – 1 yx + y = 2x – 1 y + 1 = 2x – yx y + 1 = (2 – y)x = xy + 1 2 – y Set f(x) = y = , clear the denominator then solve for x. x + 1 2 – x
Inverse Functions b. Show that f(f –1(x)) = x.
Inverse Functions 2x – 1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) =
Inverse Functions 2x – 1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) =
Inverse Functions 2x – 1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x
Inverse Functions 2x – 1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( x + 1 2 – x + 1
Inverse Functions 2x – 1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( (2 – x) (2 – x) clear denominator [ [ ] ]x + 1 2 – x + 1
Inverse Functions 2x – 1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( (2 – x) (2 – x) clear denominator [ [ ] ]x + 1 2 – x + 1
Inverse Functions 2x – 1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( (2 – x) (2 – x) clear denominator [ [ ] ]x + 1 2 – x + 1
Inverse Functions 2x – 1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( (2 – x) (2 – x) clear denominator [ [ ] ] (2 – x) x + 1 2 – x + 1 (2 – x)
Inverse Functions 2x – 1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( (2 – x) (2 – x) clear denominator [ [ ] ] (2 – x) = 2(x + 1) – (2 – x) (x + 1) + (2 – x) x + 1 2 – x + 1 (2 – x)
Inverse Functions 2x – 1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( (2 – x) (2 – x) clear denominator [ [ ] ] (2 – x) = 2(x + 1) – (2 – x) (x + 1) + (2 – x) = 2x + 2 – 2 + x x + 1 + 2 – x x + 1 2 – x + 1 (2 – x)
Inverse Functions 2x – 1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( (2 – x) (2 – x) clear denominator [ [ ] ] (2 – x) = 2(x + 1) – (2 – x) (x + 1) + (2 – x) = 2x + 2 – 2 + x x + 1 + 2 – x x + 1 2 – x + 1 = 3x 3 = x (2 – x)
Inverse Functions 2x – 1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( (2 – x) (2 – x) clear denominator [ [ ] ] (2 – x) = 2(x + 1) – (2 – x) (x + 1) + (2 – x) = 2x + 2 – 2 + x x + 1 + 2 – x x + 1 2 – x + 1 = 3x 3 = x Your turn: verify that f–1 (f (x)) = x. (2 – x)
Inverse Functions Exercise. Find the inverse functions of the given functions and verify the function–compositions. 1. f(x) = 2x + 3; f–1(f(x)) = x 2. f(x) = –3x + 5; f (f–1(x)) = x 3. f(x) = –x + 3; f (f–1 (x)) = x 4. f(x) = –3x – 4; f –1(f (x)) = x 5. f(x) = x + 3; f–1(f(x)) = x 6. f(x) = x + 5; f (f–1(x)) = x 7. f(x) = + ; f–1 (f(x)) = x 8. f(x) = x – ; f –1(f (x)) = x 1 2 –2 3 –x 2 2 3 –3 4 1 3 9. f(x) = ; f–1(f(x)) = x3 2x 10. f(x) = ; f (f–1(x)) = x–1 2 + x 11. f(x) = ; f (f–1(x)) = x 3 2 – x 12. f(x) = ; f–1(f(x)) = x3 2x + 1 13. f(x) = ; f–1(f(x)) = x 14. f(x) = ; f (f–1(x)) = xx – 1 2 + x 15. f(x) = ; f (f–1 (x)) = xx – 1 2x – 1 16. f(x) = ; f–1(f(x)) = x3 – x 2x + 3 x + 3 2 – x 17. f(x) = 1 + x3; f (f–1 (x)) = x 18. f(x) = 3x3 – 2 ; f (f–1 (x)) = x 19. f(x) = 1 + x1/3; f (f–1 (x)) = x 20. f(x) = 3x1/3 – 2 ; f (f–1 (x)) = x

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1.6 inverse function (optional)

  • 1.
  • 2.
    Inverse Functions A functionf(x) = y takes an input x and produces one output y.
  • 3.
    Inverse Functions A functionf(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x?
  • 4.
    Inverse Functions A functionf(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x? This procedure of associating the output y to the input x may or may not be a function.
  • 5.
    Inverse Functions A functionf(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x? This procedure of associating the output y to the input x may or may not be a function. If it is a function, it is called the inverse function of f(x)
  • 6.
    Inverse Functions A functionf(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x? This procedure of associating the output y to the input x may or may not be a function. If it is a function, it is called the inverse function of f(x) and it is denoted as f -1(y).
  • 7.
    Inverse Functions A functionf(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x? This procedure of associating the output y to the input x may or may not be a function. If it is a function, it is called the inverse function of f(x) and it is denoted as f -1(y). We say f(x) and f -1(y) are the inverse of each other.
  • 8.
    Inverse Functions A functionf(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x? This procedure of associating the output y to the input x may or may not be a function. If it is a function, it is called the inverse function of f(x) and it is denoted as f -1(y). We say f(x) and f -1(y) are the inverse of each other. Example A. Let f(x) = x2 = y. Suppose y = 9, what is (are) the input x that produces y = 9? Is this reverse procedure a function?
  • 9.
    Inverse Functions A functionf(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x? This procedure of associating the output y to the input x may or may not be a function. If it is a function, it is called the inverse function of f(x) and it is denoted as f -1(y). We say f(x) and f -1(y) are the inverse of each other. Example A. Let f(x) = x2 = y. Suppose y = 9, what is (are) the input x that produces y = 9? Is this reverse procedure a function? Since f(x) = x2 = 9,
  • 10.
    Inverse Functions A functionf(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x? This procedure of associating the output y to the input x may or may not be a function. If it is a function, it is called the inverse function of f(x) and it is denoted as f -1(y). We say f(x) and f -1(y) are the inverse of each other. Example A. Let f(x) = x2 = y. Suppose y = 9, what is (are) the input x that produces y = 9? Is this reverse procedure a function? Since f(x) = x2 = 9, so x = ±√9 x = – 3, x = 3.
  • 11.
    Inverse Functions A functionf(x) = y takes an input x and produces one output y. We like to do the reverse, that is, if we know the output y, what was (were) the input x? This procedure of associating the output y to the input x may or may not be a function. If it is a function, it is called the inverse function of f(x) and it is denoted as f -1(y). We say f(x) and f -1(y) are the inverse of each other. Example A. Let f(x) = x2 = y. Suppose y = 9, what is (are) the input x that produces y = 9? Is this reverse procedure a function? Since f(x) = x2 = 9, so x = ±√9 x = – 3, x = 3. This reverse procedure takes y = 9 and associates to it two different answers so it is not a function. What condition is needed for a function to have an inverse?
  • 12.
    Inverse Functions The reverseof the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs
  • 13.
    Inverse Functions The reverseof the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9).
  • 14.
    Inverse Functions The reverseof the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9).
  • 15.
    A function isone-to-one if different inputs produce different outputs. Inverse Functions The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9).
  • 16.
    Inverse Functions The reverseof the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v).
  • 17.
    A function isone-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, Inverse Functions a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9).
  • 18.
    Inverse Functions u v a one-to-onefunction The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). any pair u = v A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures,
  • 19.
    Inverse Functions u f(u) v aone-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
  • 20.
    Inverse Functions u f(u) vf(v) f(u) = f(v) a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
  • 21.
    Inverse Functions u f(u) vf(v) f(u) = f(v) a one-to-one function not a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
  • 22.
    Inverse Functions u f(u) vf(v) f(u) = f(v) a one-to-one function u v not a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). there exist u = v A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
  • 23.
    Inverse Functions u f(u) vf(v) f(u) = f(v) a one-to-one function u v not a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). there exist u = v A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
  • 24.
    Inverse Functions u f(u) vf(v) f(u) = f(v) a one-to-one function u v not a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). there exist u = v such that f(u)=f(v) A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
  • 25.
    Example B. a. g(x)= 2x is one-to-one Inverse Functions u f(u) v f(v) f(u) = f(v) a one-to-one function u v not a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). there exist u = v such that f(u)=f(v) A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
  • 26.
    Example B. a. g(x)= 2x is one-to-one because if u  v, then 2u  2v. Inverse Functions u f(u) v f(v) f(u) = f(v) a one-to-one function u v not a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). there exist u = v such that f(u)=f(v) A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
  • 27.
    Example B. a. g(x)= 2x is one-to-one because if u  v, then 2u  2v. b. f(x) = x2 is not one-to-one because for example 3  –3, but f(3) = f(–3) = 9. Inverse Functions u f(u) v f(v) f(u) = f(v) a one-to-one function u v not a one-to-one function The reverse of the function f(x) = x2 fails to be a function because x2 produces the same output with two or more different inputs (e.g. f(3) = f(–3) = 9). This prevents us from knowing exactly what x is (given that we know the output is 9). there exist u = v such that f(u)=f(v) A function is one-to-one if different inputs produce different outputs. That is, f(x) is said to be one-to-one if for every two different inputs u and v then f(u)  f(v). In pictures, any pair u = v
  • 28.
    Fact: If y= f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. Inverse Functions
  • 29.
    Fact: If y= f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function f(x) u
  • 30.
    Fact: If y= f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function u f(u) v f(v) u = v f(u) = f(v) f –1(y) is a well defined function f –1(y)f(x) u
  • 31.
    Fact: If y= f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions Given a simple y = f(x) we may solve equation y = f(x) for x in terms of y to find f –1(y) explicitly. f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function u f(u) v f(v) u = v f(u) = f(v) f –1(y) is a well defined function f –1(y)f(x) u
  • 32.
    Fact: If y= f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions Given a simple y = f(x) we may solve equation y = f(x) for x in terms of y to find f –1(y) explicitly. Example C. Find the inverse function of y = f(x) = x – 5 3 4 f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function u f(u) v f(v) u = v f(u) = f(v) f –1(y) is a well defined function f –1(y)f(x) u
  • 33.
    Fact: If y= f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions Given a simple y = f(x) we may solve equation y = f(x) for x in terms of y to find f –1(y) explicitly. Example C. Find the inverse function of y = f(x) = x – 5 Given y = x – 5, clear the denominator to solve for x.3 4 3 4 f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function u f(u) v f(v) u = v f(u) = f(v) f –1(y) is a well defined function f –1(y)f(x) u
  • 34.
    Fact: If y= f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions Given a simple y = f(x) we may solve equation y = f(x) for x in terms of y to find f –1(y) explicitly. Example C. Find the inverse function of y = f(x) = x – 5 Given y = x – 5, clear the denominator to solve for x. 4y = 3x – 20 3 4 3 4 f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function u f(u) v f(v) u = v f(u) = f(v) f –1(y) is a well defined function f –1(y)f(x) u
  • 35.
    Fact: If y= f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions Given a simple y = f(x) we may solve equation y = f(x) for x in terms of y to find f –1(y) explicitly. Example C. Find the inverse function of y = f(x) = x – 5 Given y = x – 5, clear the denominator to solve for x. 4y = 3x – 20 4y + 20 = 3x 3 4 3 4 f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function u f(u) v f(v) u = v f(u) = f(v) f –1(y) is a well defined function f –1(y)f(x) u
  • 36.
    Fact: If y= f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions Given a simple y = f(x) we may solve equation y = f(x) for x in terms of y to find f –1(y) explicitly. Example C. Find the inverse function of y = f(x) = x – 5 Given y = x – 5, clear the denominator to solve for x. 4y = 3x – 20 4y + 20 = 3x = x 3 4 4y + 20 3 3 4 f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function u f(u) v f(v) u = v f(u) = f(v) f –1(y) is a well defined function f –1(y)f(x) u
  • 37.
    Fact: If y= f(x) is one-to-one, then f –1(y) exists, i.e. the reverse procedure for f(x) is a function. In picture, Inverse Functions Given a simple y = f(x) we may solve equation y = f(x) for x in terms of y to find f –1(y) explicitly. Example C. Find the inverse function of y = f(x) = x – 5 Given y = x – 5, clear the denominator to solve for x. 4y = 3x – 20 4y + 20 = 3x = x 3 4 4y + 20 3 3 4 f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function u f(u) v f(v) u = v f(u) = f(v) f –1(y) is a well defined function f –1(y)f(x) u (Note: In general it’s impossible to solve for x explicitly.)
  • 38.
    Inverse Functions Let fand f –1 be a pair of inverse functions and that f(a) = b
  • 39.
    Inverse Functions Let fand f –1 be a pair of inverse functions and that f(a) = b ba f(a) = b
  • 40.
    Inverse Functions Let fand f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). ba f(a) = b a = f –1(b)
  • 41.
    Inverse Functions Theorem: Iff(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b).
  • 42.
    Inverse Functions Theorem: Iff(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). f(x)x f(x)
  • 43.
    Inverse Functions Theorem: Iff(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). f(x) f –1(f(x)) = x x f(x)
  • 44.
    Inverse Functions Theorem: Iff(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). f(x) f –1(f(x)) = x x f(x)
  • 45.
    Inverse Functions Theorem: Iff(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). f(x) f –1(f(x)) = x x f(x) y f –1(y) f –1(y)
  • 46.
    Inverse Functions Theorem: Iff(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
  • 47.
    Inverse Functions Theorem: Iff(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). Example D. Given the pair of inverse functions f(x) = x – 5 and show that f(f –1(y)) = y. 3 4 f –1(y) = 4y + 20 3 f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
  • 48.
    Inverse Functions Theorem: Iff(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). Example D. Given the pair of inverse functions f(x) = x – 5 and show that f(f –1(y)) = y. f (f –1(y)) = 3 4 f –1(y) = 4y + 20 3 f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
  • 49.
    Inverse Functions Theorem: Iff(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). Example D. Given the pair of inverse functions f(x) = x – 5 and show that f(f –1(y)) = y. f (f –1(y)) = f ( ) 3 4 f –1(y) = 4y + 20 3 4y + 20 3 f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
  • 50.
    Inverse Functions Theorem: Iff(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). Example D. Given the pair of inverse functions f(x) = x – 5 and show that f(f –1(y)) = y. f (f –1(y)) = f ( ) = ( ) – 5 3 4 f –1(y) = 4y + 20 3 4y + 20 3 4y + 20 3 3 4 f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
  • 51.
    Inverse Functions Theorem: Iff(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). Example D. Given the pair of inverse functions f(x) = x – 5 and show that f(f –1(y)) = y. f (f –1(y)) = f ( ) = ( ) – 5 3 4 f –1(y) = 4y + 20 3 4y + 20 3 4y + 20 3 3 4 f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
  • 52.
    Inverse Functions Theorem: Iff(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). Example D. Given the pair of inverse functions f(x) = x – 5 and show that f(f –1(y)) = y. f (f –1(y)) = f ( ) = ( ) – 5 = – 5 3 4 f –1(y) = 4y + 20 3 4y + 20 3 4y + 20 3 4y + 20 4 3 4 f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
  • 53.
    Inverse Functions Theorem: Iff(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). Example D. Given the pair of inverse functions f(x) = x – 5 and show that f(f –1(y)) = y. f (f –1(y)) = f ( ) = ( ) – 5 = – 5 3 4 f –1(y) = 4y + 20 3 4y + 20 3 4y + 20 3 4y + 20 4 4(y + 5) 4 = – 5 3 4 f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
  • 54.
    Inverse Functions Theorem: Iff(x) and f -1(y) are the inverse of each other, then f –1(f(x)) = x and that f(f –1 (y)) = y. Let f and f –1 be a pair of inverse functions and that f(a) = b then a = f –1(b). Example D. Given the pair of inverse functions f(x) = x – 5 and show that f(f –1(y)) = y. f (f –1(y)) = f ( ) = ( ) – 5 = – 5 3 4 f –1(y) = 4y + 20 3 4y + 20 3 4y + 20 3 4y + 20 4 4(y + 5) 4 = – 5 = y + 5 – 5 = y 3 4 f(x) f –1(f(x)) = x x f(x) f –1(y) f(f–1 (y) = y f –1(y) y
  • 55.
    Inverse Functions Since weusually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3
  • 56.
    Inverse Functions Since weusually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = .
  • 57.
    Inverse Functions Since weusually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = . 2x – 1 x + 1 Set f(x) = y = ,
  • 58.
    Inverse Functions Since weusually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = . 2x – 1 x + 1 Set f(x) = y = , clear the denominator then solve for x.
  • 59.
    Inverse Functions Since weusually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = . 2x – 1 x + 1 y(x + 1) = 2x – 1 Set f(x) = y = , clear the denominator then solve for x.
  • 60.
    Inverse Functions Since weusually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = . 2x – 1 x + 1 y(x + 1) = 2x – 1 yx + y = 2x – 1 Set f(x) = y = , clear the denominator then solve for x.
  • 61.
    Inverse Functions Since weusually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = . 2x – 1 x + 1 y(x + 1) = 2x – 1 yx + y = 2x – 1 y + 1 = 2x – yx Set f(x) = y = , clear the denominator then solve for x.
  • 62.
    Inverse Functions Since weusually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = . 2x – 1 x + 1 y(x + 1) = 2x – 1 yx + y = 2x – 1 y + 1 = 2x – yx y + 1 = (2 – y)x Set f(x) = y = , clear the denominator then solve for x.
  • 63.
    Inverse Functions Since weusually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = . 2x – 1 x + 1 y(x + 1) = 2x – 1 yx + y = 2x – 1 y + 1 = 2x – yx y + 1 = (2 – y)x = xy + 1 2 – y Set f(x) = y = , clear the denominator then solve for x.
  • 64.
    Inverse Functions Since weusually use x as the independent variable for functions so we often write the inverse as f –1(x) such as f–1(x) = .4x + 20 3 Hence f–1(x) = 2x – 1 x + 1 Example D. a. Find the inverse functions f–1(x) of f(x) = . 2x – 1 x + 1 y(x + 1) = 2x – 1 yx + y = 2x – 1 y + 1 = 2x – yx y + 1 = (2 – y)x = xy + 1 2 – y Set f(x) = y = , clear the denominator then solve for x. x + 1 2 – x
  • 65.
    Inverse Functions b. Showthat f(f –1(x)) = x.
  • 66.
    Inverse Functions 2x –1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) =
  • 67.
    Inverse Functions 2x –1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) =
  • 68.
    Inverse Functions 2x –1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x
  • 69.
    Inverse Functions 2x –1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( x + 1 2 – x + 1
  • 70.
    Inverse Functions 2x –1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( (2 – x) (2 – x) clear denominator [ [ ] ]x + 1 2 – x + 1
  • 71.
    Inverse Functions 2x –1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( (2 – x) (2 – x) clear denominator [ [ ] ]x + 1 2 – x + 1
  • 72.
    Inverse Functions 2x –1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( (2 – x) (2 – x) clear denominator [ [ ] ]x + 1 2 – x + 1
  • 73.
    Inverse Functions 2x –1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( (2 – x) (2 – x) clear denominator [ [ ] ] (2 – x) x + 1 2 – x + 1 (2 – x)
  • 74.
    Inverse Functions 2x –1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( (2 – x) (2 – x) clear denominator [ [ ] ] (2 – x) = 2(x + 1) – (2 – x) (x + 1) + (2 – x) x + 1 2 – x + 1 (2 – x)
  • 75.
    Inverse Functions 2x –1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( (2 – x) (2 – x) clear denominator [ [ ] ] (2 – x) = 2(x + 1) – (2 – x) (x + 1) + (2 – x) = 2x + 2 – 2 + x x + 1 + 2 – x x + 1 2 – x + 1 (2 – x)
  • 76.
    Inverse Functions 2x –1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( (2 – x) (2 – x) clear denominator [ [ ] ] (2 – x) = 2(x + 1) – (2 – x) (x + 1) + (2 – x) = 2x + 2 – 2 + x x + 1 + 2 – x x + 1 2 – x + 1 = 3x 3 = x (2 – x)
  • 77.
    Inverse Functions 2x –1 x + 1 b. Show that f(f –1(x)) = x. We have and that f–1(x) = x + 1 2 – xf (x) = Hence that f(f –1(x)) = f( )x + 1 2 – x = x + 1 2 – x – 12 )( )( (2 – x) (2 – x) clear denominator [ [ ] ] (2 – x) = 2(x + 1) – (2 – x) (x + 1) + (2 – x) = 2x + 2 – 2 + x x + 1 + 2 – x x + 1 2 – x + 1 = 3x 3 = x Your turn: verify that f–1 (f (x)) = x. (2 – x)
  • 78.
    Inverse Functions Exercise. Findthe inverse functions of the given functions and verify the function–compositions. 1. f(x) = 2x + 3; f–1(f(x)) = x 2. f(x) = –3x + 5; f (f–1(x)) = x 3. f(x) = –x + 3; f (f–1 (x)) = x 4. f(x) = –3x – 4; f –1(f (x)) = x 5. f(x) = x + 3; f–1(f(x)) = x 6. f(x) = x + 5; f (f–1(x)) = x 7. f(x) = + ; f–1 (f(x)) = x 8. f(x) = x – ; f –1(f (x)) = x 1 2 –2 3 –x 2 2 3 –3 4 1 3 9. f(x) = ; f–1(f(x)) = x3 2x 10. f(x) = ; f (f–1(x)) = x–1 2 + x 11. f(x) = ; f (f–1(x)) = x 3 2 – x 12. f(x) = ; f–1(f(x)) = x3 2x + 1 13. f(x) = ; f–1(f(x)) = x 14. f(x) = ; f (f–1(x)) = xx – 1 2 + x 15. f(x) = ; f (f–1 (x)) = xx – 1 2x – 1 16. f(x) = ; f–1(f(x)) = x3 – x 2x + 3 x + 3 2 – x 17. f(x) = 1 + x3; f (f–1 (x)) = x 18. f(x) = 3x3 – 2 ; f (f–1 (x)) = x 19. f(x) = 1 + x1/3; f (f–1 (x)) = x 20. f(x) = 3x1/3 – 2 ; f (f–1 (x)) = x