Answer
$$y=x^2$$
The graphs of $f(x)$ (red) and $f'(x)$ (green) are plotted below.
Work Step by Step
$$y=\frac{x^3}{3}$$
a) Find the derivative:
$$y'=\lim_{z\to x}\frac{f(z)-f(x)}{z-x}=\lim_{z\to x}\frac{\frac{z^3}{3}-\frac{x^3}{3}}{z-x}=\lim_{z\to x}\frac{\frac{z^3-x^3}{3}}{z-x}$$ $$y'=\lim_{z\to x}\frac{(z-x)(z^2+zx+x^2)}{3(z-x)}=\lim_{z\to x}\frac{z^2+zx+x^2}{3}$$ $$y'=\frac{x^2+x\times x+x^2}{3}=\frac{3x^2}{3}=x^2$$
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 $(-\infty,0)\cup(0,\infty)$, $f'(x)$ is positive.
- At $x=0$, $f'(x)$ is zero.
- There is no interval where $f'(x)$ is negative.
d) From the left graph, we see that on $(-\infty,0)\cup(0,\infty)$, $f(x)$ increases as $x$ increases. This corresponds to the interval where $f'$ is positive.
