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.\n\n$f(x)=x^{1 / 3}left(x^{2}-4\right)$\n\nfind 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.\n\n(simplify your answers. type exact answers, using radicals as needed.)\n\na. 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 $f(square)=square$\n\nb. the function has a local minimum value at one value of $x$. the minimum value is $f(square)=square$\n\nc. 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 $f(square)=square$\n\nd. there are no local minima.

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.\n\n$f(x)=x^{1 / 3}left(x^{2}-4\right)$\n\nfind 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.\n\n(simplify your answers. type exact answers, using radicals as needed.)\n\na. 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 $f(square)=square$\n\nb. the function has a local minimum value at one value of $x$. the minimum value is $f(square)=square$\n\nc. 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 $f(square)=square$\n\nd. there are no local minima.

Answer

Explanation:

Step1: Find the derivative

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}(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 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). Then (x^{2}=\frac{4}{7}), (x=\pm\frac{2}{\sqrt{7}}=\pm\frac{2\sqrt{7}}{7}). The derivative is undefined at (x = 0), but (f(x)) is defined at (x = 0).

Step3: Test intervals

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

Since the function changes from increasing to decreasing at (x =-\frac{2\sqrt{7}}{7}) (local maximum) and from decreasing to increasing at (x=\frac{2\sqrt{7}}{7})

Step4: Calculate local minimum value

[ 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}\times(-\frac{24}{7}) ] [ \begin{align*} f(\frac{2\sqrt{7}}{7})&=\frac{2^{1/3}\times7^{1/6}}{7^{1/3}}\times(-\frac{24}{7})\ &=2^{1/3}\times7^{-1/6}\times(-\frac{24}{7})\ &=-\frac{24\times2^{1/3}}{7^{7/6}} \end{align*} ]

Answer:

B. The function has a local minimum value at one value of (x). The minimum value is (f(\frac{2\sqrt{7}}{7})) (after simplification of the expression calculated above)