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The Algebric Functions

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The Algebric Functions

1. 1. The Algebra of Functions T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com
2. 2. Using basic algebraic functions, what limitations are there when working with real numbers? A) You CANNOT divide by zero.  Any values that would result in a zero denominator are NOT allowed, therefore the domain of the function (possible x values) would be limited. B) You CANNOT take the square root (or any even root) of a negative number.  Any values that would result in negatives under an even radical (such as square roots) result in a domain restriction.
3. 3. Example  Find the domain:  There is an x under an even radical AND x terms in the denominator, so we must consider both of these as possible limitations to our domain. 65 2 2 xx x }3,2:{: 3,2,0)2)(3( 065 2,02 2 xxxDomain xxx xx xx
4. 4. Find the indicated function values and determine whether the given values are in the domain of the function. f(1) and f(5), for f(1) = Since f(1) is defined, 1 is in the domain of f. f(5) = Since division by 0 is not defined, the number 5 is not in the domain of f. 1 1 1 1 5 4 4 1 1 5 5 0 1 ( ) 5 f x x
5. 5. Find the domain of the function Solution: We can substitute any real number in the numerator, but we must avoid inputs that make the denominator 0. Solve, x2 3x 28 = 0. (x 7)(x + 4) = 0 x 7 = 0 or x + 4 = 0 x = 7 or x = 4 2 2 3 10 8 ( ) 3 28 x x g x x x The domain consists of the set of all real numbers except x= 4 and x= 7 or {x | x 4 and x 7}. , 4 ( 4,7) (7, )
6. 6. Rational Functions  To find the domain of a function that has a variable in the denominator, set the denominator equal to zero and solve the equation. All solutions to that equation are then removed from consideration for the domain. Find the domain:  Since the radical is defined only for values that are greater than or equal to zero, solve the inequality ( ) 5f x x 5 0x 5x 5x ( ,5]
7. 7. Visualizing Domain and Range Keep the following in mind regarding the graph of a function:  Domain = the set of a function’s inputs; found on the x-axis (horizontal).  The domain of a function is the set of all “first coordinates” of the ordered pairs of a relation  Range = the set of a function’s outputs; found on the y-axis (vertical).  The range of a function is the set of all “second coordinates” of the ordered pairs of a relation.
8. 8. Example Graph the function. Then estimate the domain and range. (Note: Square root function moved one unit right) Domain = [1, ) Range = [0, ) ( ) 1f x x ( ) 1f x x
9. 9. Algebra of functions  (f + g)(x) = f(x) + g(x)  (f - g)(x) = f(x) – g(x)  (fg)(x) = f(x)g(x) 0)(, )( )( )( xg xg xf x g f
10. 10. Example Find each function and state its domain:  f + g  f – g  f ·g  f /g ;1 1g xf x x x ;1 : 11x Domainf xx xg x ;1 : 11x Domainf xx xg x 2 1; :1 1 1x x Domaing xx xf x 1 ; : 1 1 x D x f omain x xg x
11. 11. BA Their sum f + g is the function given by (f + g)(x) = f(x) + g(x) The domain of f + g consists of the numbers x that are in the domain of f and in the domain of g. Their difference f - g is the function given by (f – g ) (x) = f(x) - g(x) The domain of f – g consists of the numbers x that are in the domain of f and in the domain of g. BA If f and g are functions with domains A and B:
12. 12. Their product f g is the function given by BA The domain of f g consists of the numbers x that are in the domain of f and in the domain of g. Their quotient f /g is the function given by (f / g ) (x) = f(x) / g(x) where g(x) ≠ 0; (f g)(x) = f(x) g(x) If f and g are functions with domains A and B: The domain of f / g consists of the numbers x for which g(x) 0 that are in the domain of f and in the domain of g. 0)(xgBA
13. 13. Example Given that f(x) = x + 2 and g(x) = 2x + 5, find each of the following. a) (f + g)(x) b) (f + g)(5) a) ( )( ) ( ) ( ) 2 2 5 3 7 f g x f x g x x x x b) We can find (f + g)(5) provided 5 is in the domain of each function. This is true. f(5) = 5 + 2 = 7 g(5) = 2(5) + 5 = 15 (f + g)(5) = f(5) + g(5) = 7 + 15 = 22 (f + g)(5) = 3(5) + 7 = 22or
14. 14. Example Given that f(x) = x + 2 and g(x) = 2x + 5, find each of the following a) (f - g)(x) b) (f - g)(5) a) ( )( ) ( ) ( ) 2 (2 5) 2 2 5 3 f g x f x g x x x x x x b) We can find (f - g)(5) provided 5 is in the domain of each function. This is true. f(5) = 5 + 2 = 7 g(5) = 2(5) + 5 = 15 (f - g)(5) = f(5) - g(5) = 7 - 15 = -8 (f - g)(5) = -(5) - 3 = -8or
15. 15. Example Given that f(x) = x + 2 and g(x) = 2x + 5, find each of the following. a) (f g)(x) b) (f g)(5) a) 2 ( )( ) ( ) ( ) ( 2)(2 5) 2 9 10 fg x f x g x x x x x b) We can find (f g)(5) provided 5 is in the domain of each function. This is true. f(5) = 5 + 2 = 7 g(5) = 2(5) + 5 = 15 (f g)(5) = f(5)g(5) = 7 (15) = 105 or (f g)(5) = 2(25) + 9(5) + 10 = 105
16. 16. ( ) ( ) f x g x 2 3 16 x x The domain of f / g is {x | x > 3, x 4}. ( ) 3f x x 2 ( ) 16g x x Given the functions below, find (f/g)(x) and give the domain. ( / )( )f g x The radicand x – 3 cannot be negative. Solving x – 3 ≥ 0 gives x ≥ 3 We must exclude x = -4 and x = 4 from the domain since g(x) = 0 when x = 4.
17. 17. Composition of functions  Composition of functions is the successive application of the functions in a specific order.  Given two functions f and g, the composite function is defined by and is read “f of g of x.”  The domain of is the set of elements x in the domain of g such that g(x) is in the domain of f.  Another way to say that is to say that “the range of function g must be in the domain of function f.”  Composition of functions means the output from the inner function becomes the input of the outer function. f g f g x f g x f g
18. 18.  Composition of functions means the output from the inner function becomes the input of the outer function.  f(g(3)) means you evaluate function g at x=3, then plug that value into function f in place of the x.  Notation for composition: ))(())(( xgfxgf  f g x g(x) f(g(x)) domain of g range of f range of g domain of f g f
19. 19. f g x f g x 1 2 f x 1 2x 1 2x . gf x xgxxf Find 2 1 )(and)(Suppose Suppose f x x( ) and g x x ( ) 1 2 . Find the domain of f g . The domain of f g consists of those x in the domain of g, thus x = -2 is not in the domain of f g . In addition, g(x) > 0, so 1 0 2x 2x The domain of f g is {x | x > -2}.
20. 20. 2 2 2 1 3 2 4x xf g x 2 2 2 2 1 2 6 9 1 2 12 18 1 3g x x x f x x x Example  Evaluate and :   f g x g f x 3f x x 2 2 1g x x  2 2 4 (you check) f g x x 2 2 12 17g f x x x You can see that function composition is not commutative. NOTE: This is not a formal proof of the statement.
21. 21. (Since a radicand can’t be negative in the set of real numbers, x must be greater than or equal to zero.) Example Find the domain of and :f g x g f x 1f x x g x x  1 : 0f g x x Domain x x  1 : 1g f x x Domain x x (Since a radicand can’t be negative in the set of real numbers, x – 1 must be greater than or equal to zero.)
22. 22. Solution to Previous Example :  Determine a function that gives the cost of producing the helmets in terms of the number of hours the assembly line is functioning on a given day. Cost C n C P t 2 75 2C t t 2 14 525 100 2 40 \$5 8C t t 2 75 2n P t t t 7 1000C n n
23. 23. 1. Suppose that and2 ( ) 1f x x ( ) 3g x x ( ) ?g f x ( ) ( ( ))g f x g f x 2 ( 1)g x 2 3( 1)x 2 3 3x 2. Suppose that and2 ( ) 1f x x ( ) 3g x x (2) ?g f (2) 2g f g f 2 (2 1)g (3)g (3)(3) 9
24. 24. The End Call us for more Information: www.iTutor.com 1-855-694-8886 Visit