# To evaluate f ( x ) = 2 x x − 1 at a number, substitute the number for x in the definition of f ( x ) The given function of the valuesare f ( a ) = 2 a a − 1 , f ( a + h ) = 2 a + 2 h a + h − 1 , f ( a + h ) − f ( a ) h = − 2 h 2 ( a + h − 1 ) ( a − 1 ) Given information: consider the function f ( x ) = 2 x x − 1 Calculation: Start by assume x is the variable of a function f ( x ) . Evaluate the values of the function f ( x ) = 2 x x − 1 for the given values of the variable x Plug the value x = a , in the function f ( x ) = 2 x x − 1 thenthe value of f ( a ) . f ( a ) = 2 a a − 1 Plug the value x = a + h , in the function f ( x ) = 2 x x − 1 then the value of f ( a + h ) . f ( a + h ) = 2 ( a + h ) a + h − 1 f ( a + h ) = 2 a + 2 h a + h − 1 f ( a + h ) − f ( a ) h = 2 a + 2 h a + h − 1 − 2 a a − 1 h Since f ( a + h ) = 2 a + 2 h a + h − 1 , f ( a ) = 2 a a − 1 = ( 2 a + 2 h ) ( a − 1 ) − 2 a ( a + h − 1 ) ( a + h − 1 ) ( a − 1 ) h = 2 a 2 + 2 a h − 2 a − 2 h − ( 2 a 2 + 2 a h + 2 a ) ( a + h − 1 ) ( a − 1 ) h = − 2 h ( a + h − 1 ) ( a − 1 ) h = − 2 h 2 ( a + h − 1 ) ( a − 1 ) Thus, f ( a ) = 2 a a − 1 , f ( a + h ) = 2 a + 2 h a + h − 1 , f ( a + h ) − f ( a ) h = − 2 h 2 ( a + h − 1 ) ( a − 1 )

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

#### Solutions

Chapter 2.1, Problem 40E
Solution

## To evaluate f(x)=2xx−1 at a number, substitute the number for x in the definition of f(x)The given function of the valuesare  f(a)=2aa−1,f(a+h)=2a+2ha+h−1,f(a+h)−f(a)h=−2h2(a+h−1)(a−1)Given information:consider the function f(x)=2xx−1Calculation:Start by assume x is the variable of a function f(x) .Evaluate the values of the function f(x)=2xx−1 for the given values of the variable xPlug the value x=a, in the function f(x)=2xx−1 thenthe value of f(a).  f(a)=2aa−1Plug the value x=a+h, in the function f(x)=2xx−1 then the value of f(a+h).  f(a+h)=2(a+h)a+h−1f(a+h)=2a+2ha+h−1  f(a+h)−f(a)h=2a+2ha+h−1−2aa−1h        Since f(a+h)=2a+2ha+h−1,f(a)=2aa−1=(2a+2h)(a−1)−2a(a+h−1)(a+h−1)(a−1)h=2a2+2ah−2a−2h−(2a2+2ah+2a)(a+h−1)(a−1)h=−2h(a+h−1)(a−1)h=−2h2(a+h−1)(a−1)Thus, f(a)=2aa−1,f(a+h)=2a+2ha+h−1,f(a+h)−f(a)h=−2h2(a+h−1)(a−1)

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