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

ind the open interval(s) on which the function is increasing and decreasing. dentify 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 $f()=$, $f()=$, and $f()=$. b. the function has a local maximum value at two values of x. in increasing order of x - value, the maximum values are $f()=$ and $f()=$. c. the function has a local maximum value at one value of x. the maximum value is $f()=$ d. there are no local maxima.

ind the open interval(s) on which the function is increasing and decreasing. dentify 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 $f()=$, $f()=$, and $f()=$. b. the function has a local maximum value at two values of x. in increasing order of x - value, the maximum values are $f()=$ and $f()=$. c. the function has a local maximum value at one value of x. the maximum value is $f()=$ d. there are no local maxima.

Answer

Explanation:

Step1: Find the derivative of the function

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime). Let (u = x^{\frac{1}{3}}) and (v=x^{2}-4). Then (u^\prime=\frac{1}{3}x^{-\frac{2}{3}}) and (v^\prime = 2x). [ \begin{align*} f^\prime(x)&=\frac{1}{3}x^{-\frac{2}{3}}(x^{2}-4)+x^{\frac{1}{3}}\times(2x)\ &=\frac{x^{2}-4}{3x^{\frac{2}{3}}}+2x^{\frac{4}{3}}\ &=\frac{x^{2}-4 + 6x^{2}}{3x^{\frac{2}{3}}}\ &=\frac{7x^{2}-4}{3x^{\frac{2}{3}}} \end{align*} ]

Step2: Find the critical points

Set (f^\prime(x)=0), so (\frac{7x^{2}-4}{3x^{\frac{2}{3}}}=0). Since the denominator (3x^{\frac{2}{3}}\neq0) for (x\neq0), we solve (7x^{2}-4 = 0), (x^{2}=\frac{4}{7}), (x=\pm\frac{2}{\sqrt{7}}=\pm\frac{2\sqrt{7}}{7}). The derivative is undefined at (x = 0).

Step3: Use the first - derivative test

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

Since the function changes from increasing ((f^\prime(x)>0)) to decreasing ((f^\prime(x)<0)) at (x =-\frac{2\sqrt{7}}{7}), and from decreasing ((f^\prime(x)<0)) to increasing ((f^\prime(x)>0)) at (x=\frac{2\sqrt{7}}{7}), and has a non - differentiable point at (x = 0) (but the function does not change from increasing to decreasing or vice - versa across (x = 0) in terms of the first - derivative test near (x = 0) (left - hand side and right - hand side of (x = 0) both have (f^\prime(x)<0) in the intervals we checked)).

Step4: Calculate the function value at the local maximum point

[ f\left(-\frac{2\sqrt{7}}{7}\right)=\left(-\frac{2\sqrt{7}}{7}\right)^{\frac{1}{3}}\left(\left(-\frac{2\sqrt{7}}{7}\right)^{2}-4\right) ] [ \begin{align*} \left(-\frac{2\sqrt{7}}{7}\right)^{2}&=\frac{4\times7}{49}=\frac{4}{7}\ f\left(-\frac{2\sqrt{7}}{7}\right)&=\left(-\frac{2\sqrt{7}}{7}\right)^{\frac{1}{3}}\left(\frac{4}{7}-4\right)\ &=\left(-\frac{2\sqrt{7}}{7}\right)^{\frac{1}{3}}\left(\frac{4 - 28}{7}\right)\ &=\left(-\frac{2\sqrt{7}}{7}\right)^{\frac{1}{3}}\times\left(-\frac{24}{7}\right)\ &=\frac{24}{7}\times\left(\frac{2\sqrt{7}}{7}\right)^{\frac{1}{3}} \end{align*} ]

Answer:

The function has a local maximum value at one value of (x). The maximum value is (f\left(-\frac{2\sqrt{7}}{7}\right)=\frac{24}{7}\left(\frac{2\sqrt{7}}{7}\right)^{\frac{1}{3}})