Chapter 2: Q. 15 (page 124)

Find the different quotients of $f (x) = - 2 x^{2} + x + 1$ ;that is,find $\frac{f (x + h) - f (h)}{h}, h \neq 0$

Short Answer

Expert verified

The difference quotient is $- 4 x + 1 - 2 h$

Step by step solution

Step 1.Given information

The given Function $f (x) = - 2 x^{2} + x + 1$

we have to find $\frac{f (x + h) - f (x)}{h}$

Step 2.Compute the given function

Substitute $(x + h)$ for x in the equation, $f (x) = - 2 x^{2} + x + 1$ to obtain $f (x + h)$

$f (x + h) = - 2 {(x + h)}^{2} + (x + h) + 1$

Now compute the function $\frac{f (x - h) - f (x)}{h}$

$\frac{f (x + h) - f (x)}{h} = \frac{(- 2 {(x + h)}^{2} + (x + h) + 1) - (- 2 x^{2} + x + 1)}{h} = \frac{(- 2 (x^{2} + 2 x h + h^{2}) + x + h + 1) - (- 2 x^{2} + x + 1)}{h} = \frac{(- 2 x^{2} - 4 x h - 2 h^{2} + x + h + 1 + 2 x^{2} - x - 1)}{h}$

Step 3.Combine the like terms and simplify

$\frac{- 2 x^{2} - 4 x h - 2 h^{2} + x + h + 1 + 2 x^{2} - x - 1}{h} = \frac{(2 - 2) x^{2} (1 - 1) x + (1 - 1) - 4 x h - 2 h^{2} + h}{h} = \frac{- 4 x h + h + - 2 h^{2}}{h} = - 4 x h + 1 - 2 h$

Therefore,the difference quotient is $- 4 x + 1 - 2 h$

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Find the different quotients of $f (x) = - 2 x^{2} + x + 1$ ;that is,find $\frac{f (x + h) - f (h)}{h}, h \neq 0$

Short Answer

Step by step solution

Step 1.Given information

Step 2.Compute the given function

Step 3.Combine the like terms and simplify

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Find the different quotients off(x)=-2x2+x+1;that is,find f(x+h)-f(h)h,h≠0

Short Answer

Step by step solution

Step 1.Given information

Step 2.Compute the given function

Step 3.Combine the like terms and simplify

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Applied Mathematics

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Study anywhere. Anytime. Across all devices.

Find the different quotients of $f (x) = - 2 x^{2} + x + 1$ ;that is,find $\frac{f (x + h) - f (h)}{h}, h \neq 0$