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) )

b. the function is never decreasing.

b. find each local maximum, 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 maximum value at one value of ( x ). the maximum value is ( f(quad)= )

b. the function has a local maximum value at two values of ( x ). in increasing order of ( x )-value, the
maximum values are ( f(quad)= ) and ( f(quad)= )

c. the function has a local maximum value at three values of ( x ). in increasing order of ( x )-value, the
maximum values are ( f(quad)= ), ( f(quad)= ), and ( f(quad)= )

d. there are no local maxima.

Explanation:

Step1: Expand the function

First, expand \(f(x)=x^{\frac{1}{3}}(x^{2}-4)=x^{\frac{7}{3}}-4x^{\frac{1}{3}}\).

Step2: Find the derivative

Using the power rule \((x^{n})^\prime = nx^{n - 1}\), the derivative \(f^\prime(x)=\frac{7}{3}x^{\frac{4}{3}}-\frac{4}{3}x^{-\frac{2}{3}}=\frac{7x^{2}-4}{3x^{\frac{2}{3}}}\).

Step3: Find critical points

Set \(f^\prime(x) = 0\), so \(7x^{2}-4=0\) (since \(x
eq0\) as \(x = 0\) makes \(f^\prime(x)\) undefined but we consider the numerator for zeros of \(f^\prime(x)\)). Solving \(7x^{2}-4=0\) gives \(x=\pm\frac{2}{\sqrt{7}}=\pm\frac{2\sqrt{7}}{7}\).

Step4: Analyze the sign of \(f^\prime(x)\)

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

So the function is increasing on \((-\infty,-\frac{2\sqrt{7}}{7})\cup(\frac{2\sqrt{7}}{7},\infty)\) and decreasing on \((-\frac{2\sqrt{7}}{7},0)\cup(0,\frac{2\sqrt{7}}{7})\).

Step5: Find local maxima and minima

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

Answer:

a. The function \(f(x)\) is increasing on \((-\infty,-\frac{2\sqrt{7}}{7})\cup(\frac{2\sqrt{7}}{7},\infty)\) and decreasing on \((-\frac{2\sqrt{7}}{7},0)\cup(0,\frac{2\sqrt{7}}{7})\).
b. The function has a local maximum value at one value of \(x\). The maximum value is \(f(-\frac{2\sqrt{7}}{7})=\frac{24}{7}(\frac{2\sqrt{7}}{7})^{\frac{1}{3}}\), so the answer is A.