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.
