## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 3 - Section 3.2 - The Derivative as a Function - Exercises - Page 127: 49

#### Answer

$$y'=-2x$$ The graphs of $y=f(x)$ and $y=f'(x)$ are shown below.

#### Work Step by Step

$$y=-x^2$$ a) Find the derivative: $$y'=\lim_{z\to x}\frac{f(z)-f(x)}{z-x}=\lim_{z\to x}\frac{-z^2-(-x^2)}{z-x}=\lim_{z\to x}\frac{x^2-z^2}{-(x-z)}$$ $$y'=\lim_{z\to x}\frac{(x-z)(x+z)}{-(x-z)}=\lim_{z\to x}-(x+z)$$ $$y'=-(x+x)=-2x$$ b) The graphs are plotted below. On the left side, the red line represents the graph of $y=f(x)$. On the right side, the green line represents the graph of $y=f'(x)$. c) From the right graph, we see that - On $(-\infty,0)$, $f'(x)$ is positive. - At $x=0$, $f'(x)$ is zero. - On $(0,\infty)$, $f'(x)$ is negative. d) From the left graph, we see that - On $(-\infty,0)$, $f(x)$ increases as $x$ increases. This corresponds to the interval where $f'$ is positive. - On $(0,\infty)$, $f(x)$ decreases as $x$ increases. This corresponds to the interval where $f'$ is negative.

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