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.\n\n$g(x)=x\\sqrt{8 - x^{2}}$\n\na. on what open interval(s), if any, is the function increasing? select the correct choice below and,\nif necessary, fill in the answer box(es) to complete your choice.\n\n○ a. the function g is increasing on the open interval(s) \n(type your answer in interval notation. type an exact answer in simplified form. use a comma to\nseparate answers as needed.)\n\n○ b. the function is never increasing.

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.\n\n$g(x)=x\\sqrt{8 - x^{2}}$\n\na. on what open interval(s), if any, is the function increasing? select the correct choice below and,\nif necessary, fill in the answer box(es) to complete your choice.\n\n○ a. the function g is increasing on the open interval(s) \n(type your answer in interval notation. type an exact answer in simplified form. use a comma to\nseparate answers as needed.)\n\n○ b. the function is never increasing.

Answer

Explanation:

Step1: Find the domain of the function

For the function (g(x)=x\sqrt{8 - x^{2}}), the expression under the square - root must be non - negative. So, (8-x^{2}\geq0), which can be factored as ((2\sqrt{2}+x)(2\sqrt{2}-x)\geq0). The solution to this inequality is (x\in[- 2\sqrt{2},2\sqrt{2}]).

Step2: Find the derivative of the function

Using the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x) and (v=\sqrt{8 - x^{2}}=(8 - x^{2})^{\frac{1}{2}}). (u^\prime=1) and (v^\prime=\frac{1}{2}(8 - x^{2})^{-\frac{1}{2}}\times(-2x)=\frac{-x}{\sqrt{8 - x^{2}}}). Then (g^\prime(x)=\sqrt{8 - x^{2}}+x\times\frac{-x}{\sqrt{8 - x^{2}}}=\frac{8 - x^{2}-x^{2}}{\sqrt{8 - x^{2}}}=\frac{8 - 2x^{2}}{\sqrt{8 - x^{2}}}).

Step3: Find the critical points

Set (g^\prime(x) = 0), then (\frac{8 - 2x^{2}}{\sqrt{8 - x^{2}}}=0). Since the denominator (\sqrt{8 - x^{2}}>0) for (x\in(-2\sqrt{2},2\sqrt{2})), we solve (8 - 2x^{2}=0). (2x^{2}=8), (x^{2} = 4), (x=\pm2).

Step4: Determine the sign of the derivative

We consider the intervals ((-2\sqrt{2},-2)), ((-2,2)) and ((2,2\sqrt{2})).

  • For (x\in(-2\sqrt{2},-2)), let (x=-3) (not in the domain, but we can test the sign of (8 - 2x^{2}) by choosing a value close to (-2) in the domain, say (x=-2.5) (not valid, better to use the formula). We can also note that for (y = 8 - 2x^{2}), when (x\in(-2\sqrt{2},-2)), (y=8 - 2x^{2}<0) (since (x^{2}>4)). So (g^\prime(x)<0) on ((-2\sqrt{2},-2)).
  • For (x\in(-2,2)), let (x = 0), then (g^\prime(0)=\frac{8-0}{\sqrt{8 - 0}}=\sqrt{8}>0). So (g^\prime(x)>0) on ((-2,2)).
  • For (x\in(2,2\sqrt{2})), let (x = 2.5) (not in the domain, better to use the formula (y = 8 - 2x^{2}), when (x\in(2,2\sqrt{2})), (y=8 - 2x^{2}<0). So (g^\prime(x)<0) on ((2,2\sqrt{2})).

Answer:

A. The function (g) is increasing on the open interval((-2,2))