Question

(a) find the derivative f(x) of the function f(x) = \frac{x^{3}}{8}.
(b) graph f(x) and f(x) side by side using separate sets of coordinate axes.
(c) for what values of x, if any, is f positive? zero? negative?
(d) over what intervals of x - values, if any, does the function y = f(x) increase as x increases? decrease as x increases? how is this related to the findings in part (c)?
(b) choose the correct answer below.

Explanation:

Step1: Apply power - rule for derivative

The power - rule states that if $y = ax^n$, then $y'=nax^{n - 1}$. For $f(x)=\frac{x^{3}}{8}=\frac{1}{8}x^{3}$, where $a=\frac{1}{8}$ and $n = 3$. So $f'(x)=3\times\frac{1}{8}x^{3 - 1}=\frac{3}{8}x^{2}$.

Step2: Analyze the sign of $f'(x)$

Set $f'(x)=\frac{3}{8}x^{2}=0$. Solving for $x$, we get $x = 0$. Since $x^{2}\geq0$ for all real $x$ and $\frac{3}{8}>0$, $f'(x)=\frac{3}{8}x^{2}\geq0$ for all $x\in R$, and $f'(x)>0$ for $x
eq0$.

Step3: Analyze the increasing and decreasing of $f(x)$

A function $y = f(x)$ is increasing when $f'(x)>0$ and decreasing when $f'(x)<0$. Since $f'(x)=\frac{3}{8}x^{2}>0$ for $x
eq0$ and $f'(0) = 0$, the function $y = f(x)$ is increasing on the intervals $(-\infty,0)\cup(0,\infty)$ and it is neither increasing nor decreasing only at the single point $x = 0$.

Step4: Analyze the graphs

The function $f(x)=\frac{x^{3}}{8}$ is a cubic function passing through the origin. Its derivative $f'(x)=\frac{3}{8}x^{2}$ is a parabola opening upwards with vertex at the origin.

Answer:

(a) $f'(x)=\frac{3}{8}x^{2}$
(b) The function $f(x)=\frac{x^{3}}{8}$ is a cubic - shaped curve passing through the origin. The derivative $f'(x)=\frac{3}{8}x^{2}$ is a parabola opening upwards with vertex at $(0,0)$. Without seeing the exact options in a proper visual format, conceptually, the graph of $y = f(x)$ has an S - shape passing through the origin and $y = f'(x)$ is a U - shaped parabola opening upwards with vertex at the origin.
(c) $f'(x)$ is positive for $x
eq0$, zero for $x = 0$, and never negative.
(d) The function $y = f(x)$ increases on the intervals $(-\infty,0)\cup(0,\infty)$. The fact that $f'(x)>0$ on $(-\infty,0)\cup(0,\infty)$ indicates that the function is increasing on these intervals, and $f'(x) = 0$ at $x = 0$ is a point where the slope of the tangent line to $y = f(x)$ is zero, but the function does not change its increasing behavior around this single point.