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.
$f(x)=x^{1 / 3}left(x^{2}-4
ight)$
find each local minimum, if there are any. select the correct choice below and, if necessary, fill in the
answer box(es) to complete your choice.
(simplify your answers. type exact answers, using radicals as needed.)
a. the function has a local minimum value at two values of $x$. in increasing order of $x$-value, the
minimum values are $f(square)=square$ and $f(square)=square$
b. the function has a local minimum value at one value of $x$. the minimum value is $f(square)=square$
c. the function has a local minimum value at three values of $x$. in increasing order of $x$-value, the
minimum values are $f(square)=square, f(square)=square$, and $f(square)=square$

Explanation:

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

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x^{\frac{1}{3}}\) and \(v=x^{2}-4\).
\(u^\prime=\frac{1}{3}x^{-\frac{2}{3}}\), \(v^\prime = 2x\)
\(f^\prime(x)=\frac{1}{3}x^{-\frac{2}{3}}(x^{2}-4)+x^{\frac{1}{3}}\times2x=\frac{x^{2}-4 + 6x^{2}}{3x^{\frac{2}{3}}}=\frac{7x^{2}-4}{3x^{\frac{2}{3}}}\)

Step2: Find critical points

Set \(f^\prime(x)=0\), then \(7x^{2}-4 = 0\) (since \(x
eq0\) for the fraction to be zero).
\(x^{2}=\frac{4}{7}\), so \(x=\pm\frac{2}{\sqrt{7}}=\pm\frac{2\sqrt{7}}{7}\)

Step3: Analyze the sign of \(f^\prime(x)\)

• For \(x<-\frac{2\sqrt{7}}{7}\), let \(x=-1\), \(f^\prime(-1)=\frac{7 - 4}{3}>0\)
• For \(-\frac{2\sqrt{7}}{7}
• For \(0
• For \(x>\frac{2\sqrt{7}}{7}\), let \(x = 1\), \(f^\prime(1)=\frac{7 - 4}{3}>0\)

Step4: Find local minima

Since the function changes from decreasing to increasing at \(x =-\frac{2\sqrt{7}}{7}\) and \(x=\frac{2\sqrt{7}}{7}\)
\(f(-\frac{2\sqrt{7}}{7})=(-\frac{2\sqrt{7}}{7})^{\frac{1}{3}}((\frac{4}{7})-4)=(-\frac{2\sqrt{7}}{7})^{\frac{1}{3}}(-\frac{24}{7})\)
\(f(\frac{2\sqrt{7}}{7})=(\frac{2\sqrt{7}}{7})^{\frac{1}{3}}((\frac{4}{7})-4)=(\frac{2\sqrt{7}}{7})^{\frac{1}{3}}(-\frac{24}{7})\)

Answer:

A. The function has a local minimum value at two values of \(x\). In increasing order of \(x -\) value, the minimum values are \(f(-\frac{2\sqrt{7}}{7})=(-\frac{2\sqrt{7}}{7})^{\frac{1}{3}}(-\frac{24}{7})\) and \(f(\frac{2\sqrt{7}}{7})=(\frac{2\sqrt{7}}{7})^{\frac{1}{3}}(-\frac{24}{7})\)