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: Find the derivative of $f(x)$

Use the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$. Given $f(x)=\frac{x^{3}}{8}=\frac{1}{8}x^{3}$, then $f'(x)=\frac{1}{8}\times3x^{2}=\frac{3x^{2}}{8}$.

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

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

Step3: Relate the sign of $f'(x)$ to the increasing - decreasing behavior of $f(x)$

If $f'(x)>0$ on an interval, then $f(x)$ is increasing on that interval. If $f'(x)<0$ on an interval, then $f(x)$ is decreasing on that interval. Since $f'(x)>0$ for $x
eq0$ and $f'(x) = 0$ at $x = 0$, the function $f(x)$ is increasing for $x<0$ and $x>0$. At $x = 0$, the function has a horizontal tangent, but it is still increasing overall.

(a) # Answer:
$f'(x)=\frac{3x^{2}}{8}$

(b) The function $f(x)=\frac{x^{3}}{8}$ is a cubic function passing through the origin. Its graph has a smooth curve. The derivative $f'(x)=\frac{3x^{2}}{8}$ is a parabola opening upwards with vertex at the origin $(0,0)$. (Graphing details are skipped here as it is mainly about the solution process).

(c) # Answer:
$f'(x)>0$ for $x
eq0$, $f'(x)=0$ for $x = 0$, and $f'(x)$ is not negative for any real $x$. So the answer is not among the given options A - D. The correct statement is: $f'(x)>0$ for $x
eq0$ and $f'(x)=0$ for $x = 0$.

(d) # Answer:
The function $y = f(x)$ is increasing for $x\in(-\infty,\infty)$. This is related to the findings in part (c) because when $f'(x)>0$ (for $x
eq0$) the function is increasing, and at $x = 0$ although $f'(x)=0$, the function does not change its increasing behavior.

Answer:

Step1: Find the derivative of $f(x)$

Use the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$. Given $f(x)=\frac{x^{3}}{8}=\frac{1}{8}x^{3}$, then $f'(x)=\frac{1}{8}\times3x^{2}=\frac{3x^{2}}{8}$.

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

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

Step3: Relate the sign of $f'(x)$ to the increasing - decreasing behavior of $f(x)$

If $f'(x)>0$ on an interval, then $f(x)$ is increasing on that interval. If $f'(x)<0$ on an interval, then $f(x)$ is decreasing on that interval. Since $f'(x)>0$ for $x
eq0$ and $f'(x) = 0$ at $x = 0$, the function $f(x)$ is increasing for $x<0$ and $x>0$. At $x = 0$, the function has a horizontal tangent, but it is still increasing overall.

(a) # Answer:
$f'(x)=\frac{3x^{2}}{8}$

(b) The function $f(x)=\frac{x^{3}}{8}$ is a cubic function passing through the origin. Its graph has a smooth curve. The derivative $f'(x)=\frac{3x^{2}}{8}$ is a parabola opening upwards with vertex at the origin $(0,0)$. (Graphing details are skipped here as it is mainly about the solution process).

(c) # Answer:
$f'(x)>0$ for $x
eq0$, $f'(x)=0$ for $x = 0$, and $f'(x)$ is not negative for any real $x$. So the answer is not among the given options A - D. The correct statement is: $f'(x)>0$ for $x
eq0$ and $f'(x)=0$ for $x = 0$.

(d) # Answer:
The function $y = f(x)$ is increasing for $x\in(-\infty,\infty)$. This is related to the findings in part (c) because when $f'(x)>0$ (for $x
eq0$) the function is increasing, and at $x = 0$ although $f'(x)=0$, the function does not change its increasing behavior.