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.
a. 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.)
b. the function is never increasing

Explanation:

Step1: Find the derivative

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 critical points

Set \(f^\prime(x)=0\), so \(\frac{7x^{2}-4}{3x^{2/3}} = 0\). Since \(3x^{2/3}
eq0\) for \(x
eq0\), we solve \(7x^{2}-4=0\). Then \(x^{2}=\frac{4}{7}\), \(x=\pm\frac{2}{\sqrt{7}}=\pm\frac{2\sqrt{7}}{7}\). The derivative is undefined at \(x = 0\), but the function is defined at \(x = 0\).

Step3: Test intervals

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

Answer:

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