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

a. find the open interval(s) on which the function is increasing and decreasing.\nb. identify the functions local and absolute extreme values, if any, saying where they occur.\nf(x)=x^{1 / 3}left(x^{2}-4\right)\nthe function is ihe function is increasing on the open interval(s) \n(simplify your answer. type your answer in interval notation. type an exact answer, using radicals\nas needed. use a comma to separate answers as needed.)\nob. the function is never increasing.\non what open interval(s), if any, is the function decreasing? select the correct choice below and,\nif necessary, fill in the answer box(es) to complete your choice.\noa. the function is decreasing on the open interval(s) \n(s implify your answer. type your answer in interval notation. type an exact answer, using radicals\nas needed. use a comma to separate answers as needed.)\nob. the function is never decreasing

a. find the open interval(s) on which the function is increasing and decreasing.\nb. identify the functions local and absolute extreme values, if any, saying where they occur.\nf(x)=x^{1 / 3}left(x^{2}-4\right)\nthe function is ihe function is increasing on the open interval(s) \n(simplify your answer. type your answer in interval notation. type an exact answer, using radicals\nas needed. use a comma to separate answers as needed.)\nob. the function is never increasing.\non what open interval(s), if any, is the function decreasing? select the correct choice below and,\nif necessary, fill in the answer box(es) to complete your choice.\noa. the function is decreasing on the open interval(s) \n(s implify your answer. type your answer in interval notation. type an exact answer, using radicals\nas needed. use a comma to separate answers as needed.)\nob. the function is never decreasing

Answer

Explanation:

Step1: Find the derivative of the function

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

Step3: Determine the sign of the derivative in intervals

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

Answer:

The function is decreasing on the open interval (\left(-\frac{2}{\sqrt{7}},0\right)\cup\left(0,\frac{2}{\sqrt{7}}\right))