Answer
a. $y'=x^3$
b. See graph.
c. $f'(x)\gt0$ when $x\gt0$, $f'(0)=0$, and $f'(x)\lt0$ when $x\lt0$.
d. See explanations.
Work Step by Step
a. Given $y=f(x)=\frac{x^4}{4}$, we have $y'=f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{(x+h)^4/4-x^4/4}{h}=\lim_{h\to0}\frac{((x+h)^2+x^2)((x+h)^2-x^2)}{4h}=\lim_{h\to0}\frac{((x+h)^2+x^2)((x+h)+x)(x+h-x)}{4h}=\lim_{h\to0}\frac{((x+h)^2+x^2)((x+h)+x)}{4}=\frac{2x^22x}{4}=x^3$
b. See graph.
c. $f'(x)\gt0$ when $x\gt0$, $f'(0)=0$, and $f'(x)\lt0$ when $x\lt0$.
d. The function $f(x)$ increases when $x$ increases for $x\gt0$, and the corresponding derivative $f'(x)$ will be positive as shown in part-c. The function $f(x)$ decreases when $x$ increases for $x\lt0$, and the corresponding derivative $f'(x)$ will be negative as shown in part-c.
