# To evaluate f ( x ) = 3 − 5 x + 4 x 2 at a number,substitute the number for x in the definition of f ( x ) The given function of the values f ( a ) = 3 − 5 a + 4 a 2 , f ( a + h ) = 3 − 5 a − 5 h + 4 a 2 + 8 a h + 4 h 2 . f ( a + h ) − f ( a ) h = − 5 + 8 a + 4 h Given information: consider the function f ( x ) = 3 − 5 x + 4 x 2 Calculation: Start by assume x is the variable of a function f ( x ) . Evaluate the values of the function f ( x ) = 3 − 5 x + 4 x 2 for the given values of the variable x Plug the value x = a , in the function f ( x ) = 3 − 5 x + 4 x 2 then the value of f ( a ) . f ( a ) = 3 − 5 a + 4 a 2 Plug the value x = a + h , in the function f ( x ) = 3 − 5 x + 4 x 2 then the value of f ( a + h ) . f ( a + h ) = 3 − 5 ( a + h ) + 4 ( a + h ) 2 ⇒ f ( a + h ) = 3 − 5 a − 5 h + 4 a 2 + 8 a h + 4 h 2 Then, f ( a + h ) − f ( a ) h = 3 − 5 a − 5 h + 4 a 2 + 8 a h + 4 h 2 − ( 3 − 5 a + 4 a 2 ) h Since f ( a + h ) = 3 − 5 a − 5 h + 4 a 2 + 8 a h + 4 h 2 , f ( a ) = 3 − 5 a + 4 a 2 = − 5 h + 8 a h + 4 h 2 h = − 5 + 8 a + 4 h Thus, f ( a ) = 3 − 5 a + 4 a 2 , f ( a + h ) = 3 − 5 a − 5 h + 4 a 2 + 8 a h + 4 h 2 . f ( a + h ) − f ( a ) h = − 5 + 8 a + 4 h

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 41E
Solution

## To evaluate f(x)=3−5x+4x2 at a number,substitute the number for x in the definition of f(x)The given function of the values  f(a)=3−5a+4a2,f(a+h)=3−5a−5h+4a2+8ah+4h2.f(a+h)−f(a)h=−5+8a+4hGiven information:consider the function f(x)=3−5x+4x2Calculation:Start by assume x is the variable of a function f(x) .Evaluate the values of the function f(x)=3−5x+4x2 for the given values of the variable xPlug the value x=a, in the function f(x)=3−5x+4x2 then the value of f(a).  f(a)=3−5a+4a2Plug the value x=a+h, in the function f(x)=3−5x+4x2 then the value of f(a+h).  f(a+h)=3−5(a+h)+4(a+h)2⇒f(a+h)=3−5a−5h+4a2+8ah+4h2Then,  f(a+h)−f(a)h=3−5a−5h+4a2+8ah+4h2−(3−5a+4a2)hSince  f(a+h)=3−5a−5h+4a2+8ah+4h2,f(a)=3−5a+4a2=−5h+8ah+4h2h=−5+8a+4hThus, f(a)=3−5a+4a2,f(a+h)=3−5a−5h+4a2+8ah+4h2.f(a+h)−f(a)h=−5+8a+4h

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