## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 2 - Nonlinear Functions - 2.1 Properties of Functions - 2.1 Exercises - Page 55: 54

#### Answer

$$f(x) =-4 x^{2}+3 x+2$$ (a) \begin{aligned} f(x+h) =-4 x^{2}-8 h x-4 h^{2}+3 x+3 h+2 \end{aligned} (b) $$\begin{split} f(x+h)-f(x) =-8 h x-4 h^{2}+3 h \end{split}$$ (c) $$\begin{split} \frac{f(x+h)-f(x)}{h} =-8 x-4 h+3 \end{split}$$

#### Work Step by Step

$$f(x) =-4 x^{2}+3 x+2$$ (a) \begin{aligned} f(x+h) &=-4(x+h)^{2}+3(x+h)+2 \\ &=-4\left(x^{2}+2 h x+h^{2}\right)+3 x+3 h+2 \\ &=-4 x^{2}-8 h x-4 h^{2}+3 x+3 h+2 \end{aligned} (b) $$\begin{split} f(x+h)-f(x) & = -4 x^{2}-8 h x-4 h^{2}+3 x+3 h+2 \\ & \qquad-\left(-4 x^{2}+3 x+2\right) \\ & =-4 x^{2}-8 h x-4 h^{2}+3 x+3 h+2 \\ & \qquad+4 x^{2}-3 x-2 \\ &=-8 h x-4 h^{2}+3 h \end{split}$$ (c) $$\begin{split} \frac{f(x+h)-f(x)}{h} & = \frac{-8 h x-4 h^{2}+3 h}{h} \\ &=\frac{h(-8 x-4 h+3)}{h} \\ &=-8 x-4 h+3 \end{split}$$

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