# To evaluate f ( x ) = x x + 1 at a number, substitute the number for x in the definition of f ( x ) The given function of the values f ( a ) = a a + 1 , f ( a + h ) = a + h a + h + 1 , f ( a + h ) − f ( a ) h = h 2 ( a + h + 1 ) ( a + 1 ) Given information: consider the function f ( x ) = x x + 1 Calculation: Start by assume x is the variable of a function f ( x ) . Evaluate the values of the function f ( x ) = x x + 1 for the given values of the variable x Plug the value x = a , in the function f ( x ) = x x + 1 then the value of f ( a ) . f ( a ) = a a + 1 Plug the value x = a + h , in the function f ( x ) = x x + 1 then the value of f ( a + h ) . f ( a + h ) = a + h a + h + 1 f ( a + h ) − f ( a ) h = a + h a + h + 1 − a a + 1 h Since f ( a + h ) = a + h a + h + 1 , f ( a ) = a a + 1 = ( a + h ) ( a + 1 ) − a ( a + h + 1 ) ( a + h + 1 ) ( a + 1 ) h = a 2 + a h + a + h − ( a 2 − a h − a ) ( a + h + 1 ) ( a + 1 ) h = h ( a + h + 1 ) ( a + 1 ) h = h 2 ( a + h + 1 ) ( a + 1 ) Thus, f ( a ) = a a + 1 , f ( a + h ) = a + h a + h + 1 , f ( a + h ) − f ( a ) h = 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 39E
Solution

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

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