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    53 inverse function (optional) 53 inverse function (optional) Presentation Transcript

    • Inverse Functions
    • Inverse FunctionsA function f(x) = y takes an input x and produces one output y.
    • Inverse FunctionsA 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, whatwas (were) the input x?
    • Inverse FunctionsA 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, whatwas (were) the input x?This procedure of associating the output y to the input x mayor may not be a function.
    • Inverse FunctionsA 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, whatwas (were) the input x?This procedure of associating the output y to the input x mayor may not be a function. If it is a function, it is called theinverse function of f(x)
    • Inverse FunctionsA 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, whatwas (were) the input x?This procedure of associating the output y to the input x mayor may not be a function. If it is a function, it is called theinverse function of f(x) and it is denoted as f -1(y).
    • Inverse FunctionsA 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, whatwas (were) the input x?This procedure of associating the output y to the input x mayor may not be a function. If it is a function, it is called theinverse 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 FunctionsA 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, whatwas (were) the input x?This procedure of associating the output y to the input x mayor may not be a function. If it is a function, it is called theinverse 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) theinput x that produces y = 9? Is this reverse procedure afunction?
    • Inverse FunctionsA 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, whatwas (were) the input x?This procedure of associating the output y to the input x mayor may not be a function. If it is a function, it is called theinverse 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) theinput x that produces y = 9? Is this reverse procedure afunction?Since f(x) = x2 = 9,
    • Inverse FunctionsA 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, whatwas (were) the input x?This procedure of associating the output y to the input x mayor may not be a function. If it is a function, it is called theinverse 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) theinput x that produces y = 9? Is this reverse procedure afunction?Since f(x) = x2 = 9,so x = ±√9x = – 3, x = 3.
    • Inverse FunctionsA 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, whatwas (were) the input x?This procedure of associating the output y to the input x mayor may not be a function. If it is a function, it is called theinverse 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) theinput x that produces y = 9? Is this reverse procedure afunction?Since f(x) = x2 = 9,so x = ±√9x = – 3, x = 3.This reverse procedure takes y = 9 and associates to it twodifferent answers so it is not a function.What condition is needed for a function to have an inverse?
    • Inverse FunctionsThe reverse of the function f(x) = x2 fails to be a functionbecause x2 produces the same output with two or moredifferent inputs
    • Inverse FunctionsThe reverse of the function f(x) = x2 fails to be a functionbecause x2 produces the same output with two or moredifferent inputs (e.g. f(3) = f(–3) = 9).
    • Inverse FunctionsThe reverse of the function f(x) = x2 fails to be a functionbecause x2 produces the same output with two or moredifferent inputs (e.g. f(3) = f(–3) = 9). This prevents us frompinpointing exactly what x is even that we know the output is 9.
    • Inverse FunctionsThe reverse of the function f(x) = x2 fails to be a functionbecause x2 produces the same output with two or moredifferent inputs (e.g. f(3) = f(–3) = 9). This prevents us frompinpointing exactly what x is even that we know the output is 9.A function is one-to-one if different inputs produce differentoutputs.
    • Inverse FunctionsThe reverse of the function f(x) = x2 fails to be a functionbecause x2 produces the same output with two or moredifferent inputs (e.g. f(3) = f(–3) = 9). This prevents us frompinpointing exactly what x is even that we know the output is 9.A function is one-to-one if different inputs produce differentoutputs. That is, f(x) is said to be one-to-one if for every twodifferent inputs u and v then f(u)  f(v).
    • Inverse FunctionsThe reverse of the function f(x) = x2 fails to be a functionbecause x2 produces the same output with two or moredifferent inputs (e.g. f(3) = f(–3) = 9). This prevents us frompinpointing exactly what x is even that we know the output is 9.A function is one-to-one if different inputs produce differentoutputs. That is, f(x) is said to be one-to-one if for every twodifferent inputs u and v then f(u)  f(v). In pictures, a one-to-one function
    • Inverse FunctionsThe reverse of the function f(x) = x2 fails to be a functionbecause x2 produces the same output with two or moredifferent inputs (e.g. f(3) = f(–3) = 9). This prevents us frompinpointing exactly what x is even that we know the output is 9.A function is one-to-one if different inputs produce differentoutputs. That is, f(x) is said to be one-to-one if for every twodifferent inputs u and v then f(u)  f(v). In pictures, u v any pair u = v a one-to-one function
    • Inverse FunctionsThe reverse of the function f(x) = x2 fails to be a functionbecause x2 produces the same output with two or moredifferent inputs (e.g. f(3) = f(–3) = 9). This prevents us frompinpointing exactly what x is even that we know the output is 9.A function is one-to-one if different inputs produce differentoutputs. That is, f(x) is said to be one-to-one if for every twodifferent inputs u and v then f(u)  f(v). In pictures, u f(u) v any pair u = v a one-to-one function
    • Inverse FunctionsThe reverse of the function f(x) = x2 fails to be a functionbecause x2 produces the same output with two or moredifferent inputs (e.g. f(3) = f(–3) = 9). This prevents us frompinpointing exactly what x is even that we know the output is 9.A function is one-to-one if different inputs produce differentoutputs. That is, f(x) is said to be one-to-one if for every twodifferent inputs u and v then f(u)  f(v). In pictures, u f(u) v f(v) any pair u = v f(u) = f(v) a one-to-one function
    • Inverse FunctionsThe reverse of the function f(x) = x2 fails to be a functionbecause x2 produces the same output with two or moredifferent inputs (e.g. f(3) = f(–3) = 9). This prevents us frompinpointing exactly what x is even that we know the output is 9.A function is one-to-one if different inputs produce differentoutputs. That is, f(x) is said to be one-to-one if for every twodifferent inputs u and v then f(u)  f(v). In pictures, u f(u) v f(v) any pair u = v f(u) = f(v) a one-to-one function a none one-to-one function
    • Inverse FunctionsThe reverse of the function f(x) = x2 fails to be a functionbecause x2 produces the same output with two or moredifferent inputs (e.g. f(3) = f(–3) = 9). This prevents us frompinpointing exactly what x is even that we know the output is 9.A function is one-to-one if different inputs produce differentoutputs. That is, f(x) is said to be one-to-one if for every twodifferent inputs u and v then f(u)  f(v). In pictures, u f(u) u v f(v) v any pair u = v f(u) = f(v) there exist u = v a one-to-one function a none one-to-one function
    • Inverse FunctionsThe reverse of the function f(x) = x2 fails to be a functionbecause x2 produces the same output with two or moredifferent inputs (e.g. f(3) = f(–3) = 9). This prevents us frompinpointing exactly what x is even that we know the output is 9.A function is one-to-one if different inputs produce differentoutputs. That is, f(x) is said to be one-to-one if for every twodifferent inputs u and v then f(u)  f(v). In pictures, u f(u) u v f(v) v any pair u = v f(u) = f(v) there exist u = v a one-to-one function a none one-to-one function
    • Inverse FunctionsThe reverse of the function f(x) = x2 fails to be a functionbecause x2 produces the same output with two or moredifferent inputs (e.g. f(3) = f(–3) = 9). This prevents us frompinpointing exactly what x is even that we know the output is 9.A function is one-to-one if different inputs produce differentoutputs. That is, f(x) is said to be one-to-one if for every twodifferent inputs u and v then f(u)  f(v). In pictures, such that u f(u) u f(u)=f(v) v f(v) v any pair u = v f(u) = f(v) there exist u = v a one-to-one function a none one-to-one function
    • Inverse FunctionsThe reverse of the function f(x) = x2 fails to be a functionbecause x2 produces the same output with two or moredifferent inputs (e.g. f(3) = f(–3) = 9). This prevents us frompinpointing exactly what x is even that we know the output is 9.A function is one-to-one if different inputs produce differentoutputs. That is, f(x) is said to be one-to-one if for every twodifferent inputs u and v then f(u)  f(v). In pictures, such that u f(u) u f(u)=f(v) v f(v) v any pair u = v f(u) = f(v) there exist u = v a one-to-one function a none one-to-one functionExample B.a. g(x) = 2x is one-to-one
    • Inverse FunctionsThe reverse of the function f(x) = x2 fails to be a functionbecause x2 produces the same output with two or moredifferent inputs (e.g. f(3) = f(–3) = 9). This prevents us frompinpointing exactly what x is even that we know the output is 9.A function is one-to-one if different inputs produce differentoutputs. That is, f(x) is said to be one-to-one if for every twodifferent inputs u and v then f(u)  f(v). In pictures, such that u f(u) u f(u)=f(v) v f(v) v any pair u = v f(u) = f(v) there exist u = v a one-to-one function a none one-to-one functionExample B.a. g(x) = 2x is one-to-onebecause if u  v, then 2u  2v.
    • Inverse FunctionsThe reverse of the function f(x) = x2 fails to be a functionbecause x2 produces the same output with two or moredifferent inputs (e.g. f(3) = f(–3) = 9). This prevents us frompinpointing exactly what x is even that we know the output is 9.A function is one-to-one if different inputs produce differentoutputs. That is, f(x) is said to be one-to-one if for every twodifferent inputs u and v then f(u)  f(v). In pictures, such that u f(u) u f(u)=f(v) v f(v) v any pair u = v f(u) = f(v) there exist u = v a one-to-one function a none one-to-one functionExample B.a. g(x) = 2x is one-to-onebecause if u  v, then 2u  2v.b. f(x) = x2 is not one-to-one because for example3  –3, but f(3) = f(–3) = 9.
    • Inverse FunctionsFact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverseprocedure for f(x) is a function.
    • Inverse FunctionsFact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverseprocedure for f(x) is a function. In picture, f(x) u f(u) v f(v) u=v f(u) = f(v) f(x) is a one-to-one function
    • Inverse FunctionsFact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverseprocedure for f(x) is a function. In picture, f(x) f –1(y) u f(u) u f(u) v f(v) v f(v) u=v f(u) = f(v) u=v f(u) = f(v) f(x) is a one-to-one function f –1(y) is a well defined function
    • Inverse FunctionsFact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverseprocedure for f(x) is a function. In picture, f(x) f –1(y) u f(u) u f(u) v f(v) v f(v) u=v f(u) = f(v) u=v f(u) = f(v) f(x) is a one-to-one function f –1(y) is a well defined functionGiven a simple y = f(x) we may solve equation y = f(x) for x interms of y to find f –1(y) explicitly.
    • Inverse FunctionsFact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverseprocedure for f(x) is a function. In picture, f(x) f –1(y) u f(u) u f(u) v f(v) v f(v) u=v f(u) = f(v) u=v f(u) = f(v) f(x) is a one-to-one function f –1(y) is a well defined functionGiven a simple y = f(x) we may solve equation y = f(x) for x interms of y to find f –1(y) explicitly.Example C. Find the inverse function of y = f(x) = 3 x – 5 4
    • Inverse FunctionsFact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverseprocedure for f(x) is a function. In picture, f(x) f –1(y) u f(u) u f(u) v f(v) v f(v) u=v f(u) = f(v) u=v f(u) = f(v) f(x) is a one-to-one function f –1(y) is a well defined functionGiven a simple y = f(x) we may solve equation y = f(x) for x interms of y to find f –1(y) explicitly.Example C. Find the inverse function of y = f(x) = 3 x – 5 4Given y = 3 x – 5, clear the denominator to solve for x. 4
    • Inverse FunctionsFact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverseprocedure for f(x) is a function. In picture, f(x) f –1(y) u f(u) u f(u) v f(v) v f(v) u=v f(u) = f(v) u=v f(u) = f(v) f(x) is a one-to-one function f –1(y) is a well defined functionGiven a simple y = f(x) we may solve equation y = f(x) for x interms of y to find f –1(y) explicitly.Example C. Find the inverse function of y = f(x) = 3 x – 5 4Given y = 3 x – 5, clear the denominator to solve for x. 4 4y = 3x – 20
    • Inverse FunctionsFact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverseprocedure for f(x) is a function. In picture, f(x) f –1(y) u f(u) u f(u) v f(v) v f(v) u=v f(u) = f(v) u=v f(u) = f(v) f(x) is a one-to-one function f –1(y) is a well defined functionGiven a simple y = f(x) we may solve equation y = f(x) for x interms of y to find f –1(y) explicitly.Example C. Find the inverse function of y = f(x) = 3 x – 5 4Given y = 3 x – 5, clear the denominator to solve for x. 4 4y = 3x – 20 4y + 20 = 3x
    • Inverse FunctionsFact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverseprocedure for f(x) is a function. In picture, f(x) f –1(y) u f(u) u f(u) v f(v) v f(v) u=v f(u) = f(v) u=v f(u) = f(v) f(x) is a one-to-one function f –1(y) is a well defined functionGiven a simple y = f(x) we may solve equation y = f(x) for x interms of y to find f –1(y) explicitly.Example C. Find the inverse function of y = f(x) = 3 x – 5 4Given y = 3 x – 5, clear the denominator to solve for x. 4 4y = 3x – 20 4y + 20 = 3x 4y + 20 = x 3
    • Inverse FunctionsFact: If y = f(x) is one-to-one, then f –1(y) exists, i.e. the reverseprocedure for f(x) is a function. In picture, f(x) f –1(y) u f(u) u f(u) v f(v) v f(v) u=v f(u) = f(v) u=v f(u) = f(v) f(x) is a one-to-one function f –1(y) is a well defined functionGiven a simple y = f(x) we may solve equation y = f(x) for x interms of y to find f –1(y) explicitly.Example C. Find the inverse function of y = f(x) = 3 x – 5 4Given y = 3 x – 5, clear the denominator to solve for x. 4 4y = 3x – 20 4y + 20 = 3x 4y + 20 = x 3(Note: In general it’s impossible to solve for x explicitly.)
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b f(a) = b a b
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b then a = f –1(b). f(a) = b a b a = f –1(b)
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b then a = f –1(b).Theorem: If f(x) and f -1(y) are the inverse of each other,then f –1(f(x)) = x
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b then a = f –1(b).Theorem: If f(x) and f -1(y) are the inverse of each other,then f –1(f(x)) = x f(x) x f(x)
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b then a = f –1(b).Theorem: If f(x) and f -1(y) are the inverse of each other,then f –1(f(x)) = x f(x) x f(x) f –1(f(x)) = x
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b then a = f –1(b).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. f(x) x f(x) f –1(f(x)) = x
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b then a = f –1(b).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. f(x) f –1(y) x f(x) f –1(y) y f –1(f(x)) = x
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b then a = f –1(b).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. f(x) f –1(y) x f(x) f –1(y) y f(f–1 (y) = y f –1(f(x)) = x
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b then a = f –1(b).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. f(x) f –1(y) x f(x) f –1(y) y f(f–1 (y) = y f –1(f(x)) = xExample D. Given the pair of inverse functions f(x) = 3 x – 5 4and f –1(y) = 4y + 20 show that f(f –1(y)) = y. 3
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b then a = f –1(b).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. f(x) f –1(y) x f(x) f –1(y) y f(f–1 (y) = y f –1(f(x)) = xExample D. Given the pair of inverse functions f(x) = 3 x – 5 4and f –1(y) = 4y + 20 show that f(f –1(y)) = y. 3f (f –1(y)) =
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b then a = f –1(b).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. f(x) f –1(y) x f(x) f –1(y) y f(f–1 (y) = y f –1(f(x)) = xExample D. Given the pair of inverse functions f(x) = 3 x – 5 4and f –1(y) = 4y + 20 show that f(f –1(y)) = y. 3f (f –1(y)) = f ( 4y + 20 ) 3
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b then a = f –1(b).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. f(x) f –1(y) x f(x) f –1(y) y f(f–1 (y) = y f –1(f(x)) = xExample D. Given the pair of inverse functions f(x) = 3 x – 5 4and f –1(y) = 4y + 20 show that f(f –1(y)) = y. 3f (f –1(y)) = f ( 4y + 20 ) 3 = 3 ( 4y + 20 ) – 5 4 3
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b then a = f –1(b).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. f(x) f –1(y) x f(x) f –1(y) y f(f–1 (y) = y f –1(f(x)) = xExample D. Given the pair of inverse functions f(x) = 3 x – 5 4and f –1(y) = 4y + 20 show that f(f –1(y)) = y. 3f (f –1(y)) = f ( 4y + 20 ) 3 = 3 ( 4y + 20 ) – 5 4 3
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b then a = f –1(b).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. f(x) f –1(y) x f(x) f –1(y) y f(f–1 (y) = y f –1(f(x)) = xExample D. Given the pair of inverse functions f(x) = 3 x – 5 4and f –1(y) = 4y + 20 show that f(f –1(y)) = y. 3f (f –1(y)) = f ( 4y + 20 ) 3 = 3 ( 4y + 20 ) – 5 4 3 = 4y + 20 – 5 4
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b then a = f –1(b).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. f(x) f –1(y) x f(x) f –1(y) y f(f–1 (y) = y f –1(f(x)) = xExample D. Given the pair of inverse functions f(x) = 3 x – 5 4and f –1(y) = 4y + 20 show that f(f –1(y)) = y. 3f (f –1(y)) = f ( 4y + 20 ) 3 = 3 ( 4y + 20 ) – 5 4 3 = 4y + 20 – 5 4 4(y + 5) = –5 4
    • Inverse FunctionsLet f and f –1 be a pair of inverse functions and thatf(a) = b then a = f –1(b).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. f(x) f –1(y) x f(x) f –1(y) y f(f–1 (y) = y f –1(f(x)) = xExample D. Given the pair of inverse functions f(x) = 3 x – 5 4and f –1(y) = 4y + 20 show that f(f –1(y)) = y. 3f (f –1(y)) = f ( 4y + 20 ) 3 = 3 ( 4y + 20 ) – 5 4 3 = 4y + 20 – 5 4 4(y + 5) = –5 =y+5–5=y 4
    • Inverse FunctionsSince we usually use x as the independent variable forfunctions, so we often write the inverse as f –1(x) such asf–1(x) = 4x + 20 . 3
    • Inverse FunctionsSince we usually use x as the independent variable forfunctions, so we often write the inverse as f –1(x) such asf–1(x) = 4x + 20 . 3Example E. 2x – 1a. Find the inverse functions f–1(x) of f(x) = x + 1 .
    • Inverse FunctionsSince we usually use x as the independent variable forfunctions, so we often write the inverse as f –1(x) such asf–1(x) = 4x + 20 . 3Example E. 2x – 1a. Find the inverse functions f–1(x) of f(x) = x + 1 . –1Set f(x) = y = 2x + 1 , x
    • Inverse FunctionsSince we usually use x as the independent variable forfunctions, so we often write the inverse as f –1(x) such asf–1(x) = 4x + 20 . 3Example E. 2x – 1a. Find the inverse functions f–1(x) of f(x) = x + 1 . –1Set f(x) = y = 2x + 1 , clear the denominator then solve for x. x
    • Inverse FunctionsSince we usually use x as the independent variable forfunctions, so we often write the inverse as f –1(x) such asf–1(x) = 4x + 20 . 3Example E. 2x – 1a. Find the inverse functions f–1(x) of f(x) = x + 1 . –1Set f(x) = y = 2x + 1 , clear the denominator then solve for x. xy(x + 1) = 2x – 1
    • Inverse FunctionsSince we usually use x as the independent variable forfunctions, so we often write the inverse as f –1(x) such asf–1(x) = 4x + 20 . 3Example E. 2x – 1a. Find the inverse functions f–1(x) of f(x) = x + 1 . –1Set f(x) = y = 2x + 1 , clear the denominator then solve for x. xy(x + 1) = 2x – 1yx + y = 2x – 1
    • Inverse FunctionsSince we usually use x as the independent variable forfunctions, so we often write the inverse as f –1(x) such asf–1(x) = 4x + 20 . 3Example E. 2x – 1a. Find the inverse functions f–1(x) of f(x) = x + 1 . –1Set f(x) = y = 2x + 1 , clear the denominator then solve for x. xy(x + 1) = 2x – 1yx + y = 2x – 1y + 1 = 2x – yx
    • Inverse FunctionsSince we usually use x as the independent variable forfunctions, so we often write the inverse as f –1(x) such asf–1(x) = 4x + 20 . 3Example E. 2x – 1a. Find the inverse functions f–1(x) of f(x) = x + 1 . –1Set f(x) = y = 2x + 1 , clear the denominator then solve for x. xy(x + 1) = 2x – 1yx + y = 2x – 1y + 1 = 2x – yxy + 1 = (2 – y)x
    • Inverse FunctionsSince we usually use x as the independent variable forfunctions, so we often write the inverse as f –1(x) such asf–1(x) = 4x + 20 . 3Example E. 2x – 1a. Find the inverse functions f–1(x) of f(x) = x + 1 . –1Set f(x) = y = 2x + 1 , clear the denominator then solve for x. xy(x + 1) = 2x – 1yx + y = 2x – 1y + 1 = 2x – yxy + 1 = (2 – y)x y+1 =x 2–y
    • Inverse FunctionsSince we usually use x as the independent variable forfunctions, so we often write the inverse as f –1(x) such asf–1(x) = 4x + 20 . 3Example E. 2x – 1a. Find the inverse functions f–1(x) of f(x) = x + 1 . –1Set f(x) = y = 2x + 1 , clear the denominator then solve for x. xy(x + 1) = 2x – 1yx + y = 2x – 1y + 1 = 2x – yxy + 1 = (2 – y)x y+1 =x 2–y x+1Hence f–1(x) = 2–x
    • Inverse Functionsb. Show that f(f –1(x)) = x.
    • Inverse Functionsb. Show that f(f –1(x)) = x.We have f (x) = 2x – 1 and that f–1(x) = 2 – x x+1 x+1
    • Inverse Functionsb. Show that f(f –1(x)) = x.We have f (x) = 2x – 1 and that f–1(x) = 2 – x x+1 x+1Hence that f(f –1(x)) =
    • Inverse Functionsb. Show that f(f –1(x)) = x.We have f (x) = 2x – 1 and that f–1(x) = 2 – x x+1 x+1Hence that f(f –1(x)) = f( x + 1 ) 2–x
    • Inverse Functionsb. Show that f(f –1(x)) = x.We have f (x) = 2x – 1 and that f–1(x) = 2 – x x+1 x+1Hence that f(f –1(x)) = f( x + 1 ) 2–x x + 1) 2(2 – x – 1 (2 – x) = 2 – (x + 1 ) (2 – x) 2–x
    • Inverse Functionsb. Show that f(f –1(x)) = x.We have f (x) = 2x – 1 and that f–1(x) = 2 – x x+1 x+1Hence that f(f –1(x)) = f( x + 1 ) 2–x x + 1) 2(2 – x – 1 (2 – x) = x + 1 ) (2 – x) clear denominator 2 –( 2–x
    • Inverse Functionsb. Show that f(f –1(x)) = x.We have f (x) = 2x – 1 and that f–1(x) = 2 – x x+1 x+1Hence that f(f –1(x)) = f( x + 1 ) 2–x x + 1) [ 2(2 – x – 1 ] (2 – x) = x+1 clear denominator [( 2 – x ) + 1 ] (2 – x)
    • Inverse Functionsb. Show that f(f –1(x)) = x.We have f (x) = 2x – 1 and that f–1(x) = 2 – x x+1 x+1Hence that f(f –1(x)) = f( x + 1 ) 2–x x + 1) [ 2(2 – x – 1 ] (2 – x) = x+1 clear denominator [( 2 – x ) + 1 ] (2 – x)
    • Inverse Functionsb. Show that f(f –1(x)) = x.We have f (x) = 2x – 1 and that f–1(x) = 2 – x x+1 x+1Hence that f(f –1(x)) = f( x + 1 ) 2 – x (2 – x) x + 1) [ 2(2 – x – 1 ] (2 – x) = x + 1 (2 – x) clear denominator [( 2 – x ) + 1 ] (2 – x)
    • Inverse Functionsb. Show that f(f –1(x)) = x.We have f (x) = 2x – 1 and that f–1(x) = 2 – x x+1 x+1Hence that f(f –1(x)) = f( x + 1 ) 2 – x (2 – x) x + 1) [ 2(2 – x – 1 ] (2 – x) = x + 1 (2 – x) clear denominator [( 2 – x ) + 1 ] (2 – x) 2(x + 1) – (2 – x) = (x + 1) + (2 – x)
    • Inverse Functionsb. Show that f(f –1(x)) = x.We have f (x) = 2x – 1 and that f–1(x) = 2 – x x+1 x+1Hence that f(f –1(x)) = f( x + 1 ) 2 – x (2 – x) x + 1) [ 2(2 – x – 1 ] (2 – x) = x + 1 (2 – x) clear denominator [( 2 – x ) + 1 ] (2 – x) 2(x + 1) – (2 – x) = (x + 1) + (2 – x) 2x + 2 – 2 + x = x+1+2–x
    • Inverse Functionsb. Show that f(f –1(x)) = x.We have f (x) = 2x – 1 and that f–1(x) = 2 – x x+1 x+1Hence that f(f –1(x)) = f( x + 1 ) 2 – x (2 – x) x + 1) [ 2(2 – x – 1 ] (2 – x) = x + 1 (2 – x) clear denominator [( 2 – x ) + 1 ] (2 – x) 2(x + 1) – (2 – x) = (x + 1) + (2 – x) 2x + 2 – 2 + x = x+1+2–x = 3x = x 3
    • Inverse Functionsb. Show that f(f –1(x)) = x.We have f (x) = 2x – 1 and that f–1(x) = 2 – x x+1 x+1Hence that f(f –1(x)) = f( x + 1 ) 2 – x (2 – x) x + 1) [ 2(2 – x – 1 ] (2 – x) = x + 1 (2 – x) clear denominator [( 2 – x ) + 1 ] (2 – x) 2(x + 1) – (2 – x) = (x + 1) + (2 – x) 2x + 2 – 2 + x = x+1+2–x = 3x = x 3Your turn: verify that f–1 (f (x)) = x.