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 )

BuyFind

Precalculus: Mathematics for Calcu...

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

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+1

Calculation:

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 x

Plug the value x=a, in the function f(x)=xx+1 then the value of f(a).

  f(a)=aa+1

Plug 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+1aa+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(a2aha)(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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