Answer
a. $y'=x^2$
b. See graph.
c. $f'(x)\gt0$ for all x except $x=0$, $f'(0)=0$, and $f'(x)$ will not be negative.
d. See explanations.
Work Step by Step
a. Given $y=f(x)=\frac{x^3}{3}$, we have $y'=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{(x+h)^3/3-x^3/3}{h}=\lim_{h\to0}\frac{x^3+3x^2h+3xh^2+h^3-x^3}{3h}=\lim_{h\to0}\frac{3x^2+3xh+h^2}{3}=x^2$
b. See graph.
c. $f'(x)\gt0$ for all the x-values in the domain except at $x=0$, $f'(0)=0$, and $f'(x)$ will not be negative.
d. The function $f(x)$ increases when $x$ increases for all the x-values in the domain except at $x=0$, and the corresponding derivative $f'(x)$ will be positive as shown in part-c. There are no regions where the function $f(x)$ decreases when $x$ increases.
