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)
if the function has extreme values, which of the extreme values, if any, are absolute? select the correct choice below and fill in any answer boxes within your choice.
(simplify your answers. type exact answers, using radicals as needed. use a comma to separate answers as needed.)
o a. there is no absolute maximum, but there is an absolute minimum of \square at x = \square.
o b. there is an absolute maximum of \square at x = \square and an absolute minimum of \square at x = \square
o c. there is an absolute maximum of \square at x = \square, but no absolute minimum.
o d. there are local extreme values but there are no absolute extreme values.
o e. there are no local or absolute extreme values.

Explanation:

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

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x^{1/3}\) and \(v=x^{2}-4\).
\(u^\prime=\frac{1}{3}x^{-2/3}\), \(v^\prime = 2x\)
\(f^\prime(x)=\frac{1}{3}x^{-2/3}(x^{2}-4)+x^{1/3}(2x)=\frac{x^{2}-4 + 6x^{2}}{3x^{2/3}}=\frac{7x^{2}-4}{3x^{2/3}}\)

Step2: Find critical points

Set \(f^\prime(x) = 0\), so \(7x^{2}-4=0\) (since \(x = 0\) makes \(f^\prime(x)\) undefined but we first solve \(7x^{2}-4=0\)).
\(x^{2}=\frac{4}{7}\), \(x=\pm\frac{2}{\sqrt{7}}=\pm\frac{2\sqrt{7}}{7}\)

Step3: Determine intervals of increase and decrease

• For \(x\in(-\infty,-\frac{2\sqrt{7}}{7})\), pick \(x=-1\), \(f^\prime(-1)=\frac{7 - 4}{3}>0\), so \(f(x)\) is increasing.
• For \(x\in(-\frac{2\sqrt{7}}{7},0)\), pick \(x =-\frac{1}{2}\), \(f^\prime(-\frac{1}{2})=\frac{7\times\frac{1}{4}-4}{3\times(\frac{1}{2})^{2/3}}<0\), so \(f(x)\) is decreasing.
• For \(x\in(0,\frac{2\sqrt{7}}{7})\), pick \(x=\frac{1}{2}\), \(f^\prime(\frac{1}{2})=\frac{7\times\frac{1}{4}-4}{3\times(\frac{1}{2})^{2/3}}<0\), so \(f(x)\) is decreasing.
• For \(x\in(\frac{2\sqrt{7}}{7},\infty)\), pick \(x = 1\), \(f^\prime(1)=\frac{7 - 4}{3}>0\), so \(f(x)\) is increasing.

Step4: Find local and absolute extrema

• Local maximum: \(f(-\frac{2\sqrt{7}}{7})=(-\frac{2\sqrt{7}}{7})^{1/3}[(-\frac{2\sqrt{7}}{7})^{2}-4]\)

\(=(-\frac{2\sqrt{7}}{7})^{1/3}(\frac{4}{7}-4)=(-\frac{2\sqrt{7}}{7})^{1/3}(-\frac{24}{7})\)

• Local minimum: \(f(\frac{2\sqrt{7}}{7})=(\frac{2\sqrt{7}}{7})^{1/3}[(\frac{2\sqrt{7}}{7})^{2}-4]=(\frac{2\sqrt{7}}{7})^{1/3}(\frac{4}{7}-4)=(\frac{2\sqrt{7}}{7})^{1/3}(-\frac{24}{7})\)

As \(x\to\pm\infty\), \(f(x)=x^{1/3}(x^{2}-4)\sim x^{7/3}\to\pm\infty\)

Answer:

D. There are local extreme values but there are no absolute extreme values.