find the open interval(s) on which the function is increasing and decreasing. identify the functions local…

find the open interval(s) on which the function is increasing and decreasing. identify the functions local and absolute extreme values, if any, saying where they occur. $f(x)=x^{1 / 3}left(x^{2}-4\right)$ simplify your answers. type exact answers, using radicals as needed.) a. the function has a local maximum value at three values of $x$. in increasing order of $x$-value, the maximum values are $f(square)=square, f(square)=square$, and $f(square)=square$ b. the function has a local maximum value at two values of $x$. in increasing order of $x$-value, the maximum values are $f(square)=square$ and $f(square)=square$ c. the function has a local maximum value at one value of $x$. the maximum value is $fleft(-\frac{2}{sqrt{7}}\right)=0$ d. there are no local maxima

find the open interval(s) on which the function is increasing and decreasing. identify the functions local and absolute extreme values, if any, saying where they occur. $f(x)=x^{1 / 3}left(x^{2}-4\right)$ simplify your answers. type exact answers, using radicals as needed.) a. the function has a local maximum value at three values of $x$. in increasing order of $x$-value, the maximum values are $f(square)=square, f(square)=square$, and $f(square)=square$ b. the function has a local maximum value at two values of $x$. in increasing order of $x$-value, the maximum values are $f(square)=square$ and $f(square)=square$ c. the function has a local maximum value at one value of $x$. the maximum value is $fleft(-\frac{2}{sqrt{7}}\right)=0$ d. there are no local maxima

Answer

Explanation:

Step1: Find the derivative of the function

First, 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). [ \begin{align*} f^\prime(x)&=\frac{1}{3}x^{-2/3}(x^{2}-4)+x^{1/3}\times(2x)\ &=\frac{x^{2}-4}{3x^{2/3}}+2x^{4/3}\ &=\frac{x^{2}-4 + 6x^{2}}{3x^{2/3}}\ &=\frac{7x^{2}-4}{3x^{2/3}} \end{align*} ]

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}\neq0) for (x\neq0), we solve (7x^{2}-4=0). [x^{2}=\frac{4}{7}\Rightarrow x=\pm\frac{2}{\sqrt{7}}] The derivative (f^\prime(x)) is undefined at (x = 0), but (x = 0) is in the domain of (f(x)).

Step3: Test the intervals

  • For (x<-\frac{2}{\sqrt{7}}), let (x=-1). Then (f^\prime(-1)=\frac{7\times(- 1)^{2}-4}{3\times(-1)^{2/3}}=\frac{3}{3}=1>0), so (f(x)) is increasing on (\left(-\infty,-\frac{2}{\sqrt{7}}\right)).
  • For (-\frac{2}{\sqrt{7}}<x<0), let (x =-\frac{1}{\sqrt{7}}). Then (f^\prime\left(-\frac{1}{\sqrt{7}}\right)=\frac{7\times\left(-\frac{1}{\sqrt{7}}\right)^{2}-4}{3\times\left(-\frac{1}{\sqrt{7}}\right)^{2/3}}=\frac{1 - 4}{3\times\left(\frac{1}{\sqrt{7}}\right)^{2/3}}<0), so (f(x)) is decreasing on (\left(-\frac{2}{\sqrt{7}},0\right)).
  • For (0<x<\frac{2}{\sqrt{7}}), let (x=\frac{1}{\sqrt{7}}). Then (f^\prime\left(\frac{1}{\sqrt{7}}\right)=\frac{7\times\left(\frac{1}{\sqrt{7}}\right)^{2}-4}{3\times\left(\frac{1}{\sqrt{7}}\right)^{2/3}}=\frac{1 - 4}{3\times\left(\frac{1}{\sqrt{7}}\right)^{2/3}}<0), so (f(x)) is decreasing on (\left(0,\frac{2}{\sqrt{7}}\right)).
  • For (x>\frac{2}{\sqrt{7}}), let (x = 1). Then (f^\prime(1)=\frac{7\times1^{2}-4}{3\times1^{2/3}}=\frac{3}{3}=1>0), so (f(x)) is increasing on (\left(\frac{2}{\sqrt{7}},+\infty\right)).

Since the function changes from increasing to decreasing at (x=-\frac{2}{\sqrt{7}}), we find (f\left(-\frac{2}{\sqrt{7}}\right)): [ \begin{align*} f\left(-\frac{2}{\sqrt{7}}\right)&=\left(-\frac{2}{\sqrt{7}}\right)^{1/3}\left(\left(-\frac{2}{\sqrt{7}}\right)^{2}-4\right)\ &=\left(-\frac{2}{\sqrt{7}}\right)^{1/3}\left(\frac{4}{7}-4\right)\ &=\left(-\frac{2}{\sqrt{7}}\right)^{1/3}\left(\frac{4 - 28}{7}\right)\ &=\left(-\frac{2}{\sqrt{7}}\right)^{1/3}\times\left(-\frac{24}{7}\right)\ \end{align*} ] [ \begin{align*} f\left(-\frac{2}{\sqrt{7}}\right)&=\left(\frac{2}{\sqrt{7}}\right)^{1/3}\times\frac{24}{7}\ \end{align*} ] The function has a local maximum at (x =-\frac{2}{\sqrt{7}})

Answer:

C. The function has a local maximum value at one value of (x). The maximum value is (f\left(-\frac{2}{\sqrt{7}}\right)) (the value of (f\left(-\frac{2}{\sqrt{7}}\right)) can be further simplified as (\frac{24}{7}\left(\frac{2}{\sqrt{7}}\right)^{1/3}))