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)? (c) determine the x - values for which f(x)>0, f(x)=0, and f(x)<0. choose the correct answer below. a. f(x)<0 for x<0 and x>0; f(x)=0 for x = 0; and it is not positive for any value of x. b. f(x)>0 for x<0 and x>0; f(x)=0 for x = 0; and it is not negative for any value of x. c. f(x)<0 for x<0; f(x)=0 for x>0; and f(x)>0 for x = 0. d. f(x)>0 for x>0; f(x)=0 for x<0; and f(x)<0 for x = 0.

Explanation:

Step1: Apply power - rule for differentiation

The power - rule states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$. For the function $f(x)=\frac{x^{3}}{8}=\frac{1}{8}x^{3}$, where $a=\frac{1}{8}$ and $n = 3$. So, $f^\prime(x)=\frac{1}{8}\times3x^{2}=\frac{3x^{2}}{8}$.

Step2: Analyze the sign of $f^\prime(x)$

Since $f^\prime(x)=\frac{3x^{2}}{8}$, and $x^{2}\geq0$ for all real $x$, and $\frac{3}{8}>0$. Then $f^\prime(x)\geq0$ for all real $x$. $f^\prime(x) = 0$ when $x = 0$, and $f^\prime(x)>0$ when $x
eq0$.

Step3: Relate the sign of $f^\prime(x)$ to the increase and decrease of $f(x)$

A function $y = f(x)$ is increasing when $f^\prime(x)>0$ and has a horizontal tangent when $f^\prime(x)=0$. Since $f^\prime(x)=\frac{3x^{2}}{8}\geq0$ for all $x\in R$, with $f^\prime(x) = 0$ at $x = 0$ and $f^\prime(x)>0$ for $x
eq0$, the function $y = f(x)$ is increasing for $x\in(-\infty,0)\cup(0,+\infty)$ and has a horizontal tangent at $x = 0$.

Answer:

(a) $f^\prime(x)=\frac{3x^{2}}{8}$
(b) To graph $f(x)=\frac{x^{3}}{8}$, it is a cubic - type function passing through the origin with a relatively flat shape near the origin. To graph $f^\prime(x)=\frac{3x^{2}}{8}$, it is a parabola opening upwards with vertex at the origin $(0,0)$.
(c) $f^\prime(x)>0$ for $x
eq0$, $f^\prime(x)=0$ for $x = 0$, and $f^\prime(x)$ is not negative for any real $x$. So the correct answer is: None of the options A - D are correct as presented. The correct statement is $f^\prime(x)>0$ for $x
eq0$ and $f^\prime(x)=0$ for $x = 0$.
(d) The function $y = f(x)$ increases for $x\in(-\infty,0)\cup(0,+\infty)$. The fact that $f^\prime(x)>0$ for $x
eq0$ indicates that the function is increasing on the intervals $(-\infty,0)$ and $(0,+\infty)$, and $f^\prime(x) = 0$ at $x = 0$ means there is a horizontal tangent at $x = 0$.