Question

a. find the open interval(s) on which the function is increasing and decreasing
b. identify the functions local and absolute extreme values, if any, saying where they occur.

f(x)=x^{1 / 3}\left(x^{2}-4\
ight)

a. on what open interval(s), if any, is the function increasing? select the correct choice below and,
if necessary, fill in the answer box(es) to complete your choice.

oa. the function is increasing on the open interval(s)
(simplify your answer. type your answer in interval notation. type an exact answer, using radicals as
needed. use a comma to separate answers as needed.)

ob. the function is never increasing.

Explanation:

Step1: Find the derivative of the function

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\). Let \(u = x^{1/3}\) and \(v=x^{2}-4\). Then \(u^\prime=\frac{1}{3}x^{-2/3}\) and \(v^\prime = 2x\).

$$
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Step2: Find the critical points

Set \(f^\prime(x)=0\), so \(\frac{7x^{2}-4}{3x^{2/3}} = 0\). Since the denominator \(3x^{2/3}
eq0\) for \(x
eq0\), we solve \(7x^{2}-4=0\).

$$ x^{2}=\frac{4}{7}\Rightarrow x=\pm\frac{2}{\sqrt{7}}=\pm\frac{2\sqrt{7}}{7} $$

The derivative \(f^\prime(x)\) is undefined at \(x = 0\) (because of the \(x^{-2/3}\) term in the first - step of derivative calculation), but the function \(f(x)\) is defined at \(x = 0\).

Step3: Test the intervals

• For the interval \((-\infty,-\frac{2\sqrt{7}}{7})\), let \(x=-1\). Then \(f^\prime(-1)=\frac{7\times(- 1)^{2}-4}{3\times(-1)^{2/3}}=\frac{7 - 4}{3}=1>0\).
• For the interval \((-\frac{2\sqrt{7}}{7},0)\), let \(x =-\frac{1}{2}\). Then \(f^\prime(-\frac{1}{2})=\frac{7\times(-\frac{1}{2})^{2}-4}{3\times(-\frac{1}{2})^{2/3}}=\frac{\frac{7}{4}-4}{3\times(-\frac{1}{2})^{2/3}}=\frac{7 - 16}{12\times(-\frac{1}{2})^{2/3}}<0\).
• For the interval \((0,\frac{2\sqrt{7}}{7})\), let \(x=\frac{1}{2}\). Then \(f^\prime(\frac{1}{2})=\frac{7\times(\frac{1}{2})^{2}-4}{3\times(\frac{1}{2})^{2/3}}=\frac{\frac{7}{4}-4}{3\times(\frac{1}{2})^{2/3}}=\frac{7 - 16}{12\times(\frac{1}{2})^{2/3}}<0\).
• For the interval \((\frac{2\sqrt{7}}{7},\infty)\), let \(x = 1\). Then \(f^\prime(1)=\frac{7\times1^{2}-4}{3\times1^{2/3}}=\frac{7 - 4}{3}=1>0\).

Answer:

A. The function is increasing on the open interval(s) \((-\infty,-\frac{2\sqrt{7}}{7}),(\frac{2\sqrt{7}}{7},\infty)\)