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.\ng(x)=x\\sqrt{18 - x^{2}}\na. the function g is increasing on the open interval(s) (-3,3)\n(type your answer in interval notation. type an exact answer in simplified form. use a comma to\nseparate answers as needed.)\nb. 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.\na. the function g is decreasing 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.)\nb. 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.\ng(x)=x\\sqrt{18 - x^{2}}\na. the function g is increasing on the open interval(s) (-3,3)\n(type your answer in interval notation. type an exact answer in simplified form. use a comma to\nseparate answers as needed.)\nb. 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.\na. the function g is decreasing 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.)\nb. the function is never decreasing

Answer

Explanation:

Step1: Find the domain of the function

For the function (g(x)=x\sqrt{18 - x^{2}}), the expression under the square - root must be non - negative. So, (18-x^{2}\geq0), which can be factored as ((\sqrt{18}+x)(\sqrt{18}-x)\geq0) or ((3\sqrt{2}+x)(3\sqrt{2}-x)\geq0). The solutions of the inequality (18 - x^{2}\geq0) are (x\in[-3\sqrt{2},3\sqrt{2}]).

Step2: Find the derivative of the function

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x) and (v=\sqrt{18 - x^{2}}=(18 - x^{2})^{\frac{1}{2}}). The derivative of (u=x) is (u^\prime = 1). The derivative of (v=(18 - x^{2})^{\frac{1}{2}}) using the chain rule: let (t = 18 - x^{2}), then (v=t^{\frac{1}{2}}), (v^\prime=\frac{1}{2}t^{-\frac{1}{2}}\cdot(-2x)=\frac{-x}{\sqrt{18 - x^{2}}}). So, (g^\prime(x)=\sqrt{18 - x^{2}}+x\cdot\frac{-x}{\sqrt{18 - x^{2}}}=\frac{18 - x^{2}-x^{2}}{\sqrt{18 - x^{2}}}=\frac{18 - 2x^{2}}{\sqrt{18 - x^{2}}}).

Step3: Find the critical points

Set (g^\prime(x)=0), then (\frac{18 - 2x^{2}}{\sqrt{18 - x^{2}}}=0). Since the denominator (\sqrt{18 - x^{2}}>0) for (x\in(-3\sqrt{2},3\sqrt{2})), we solve (18 - 2x^{2}=0). (2x^{2}=18), (x^{2}=9), (x=\pm3).

Step4: Determine the intervals of increase and decrease

We use test points in the intervals ((-3\sqrt{2},-3)), ((-3,3)), and ((3,3\sqrt{2})).

  • For the interval ((-3\sqrt{2},-3)), let (x=-4) (but (x=-4\notin[-3\sqrt{2},3\sqrt{2}])), let's take (x = - 4) (not in the domain, wrong approach). Let's use the formula (g^\prime(x)=\frac{18 - 2x^{2}}{\sqrt{18 - x^{2}}}). Take a test point (x=-3) (left - hand limit). As (x) approaches (-3) from the left ((x=-3 - h), (h>0) small), (g^\prime(x)=\frac{18-2x^{2}}{\sqrt{18 - x^{2}}}), when (x=-3 - h), (18-2x^{2}=18-2(9 + 6h+h^{2})=-12h-2h^{2}<0). Take a test point (x = 0) (in the interval ((-3,3))), (g^\prime(0)=\frac{18-0}{\sqrt{18-0}}=\sqrt{18}>0). Take a test point (x = 4) (not in the domain, wrong approach). As (x) approaches (3) from the right ((x = 3+h), (h>0) small), (18-2x^{2}=18 - 2(9 + 6h+h^{2})=-12h-2h^{2}<0).

The function (g(x)) is increasing when (g^\prime(x)>0). Solving (\frac{18 - 2x^{2}}{\sqrt{18 - x^{2}}}>0), since (\sqrt{18 - x^{2}}>0) for (x\in(-3\sqrt{2},3\sqrt{2})), we solve (18 - 2x^{2}>0), (x^{2}<9), (x\in(-3,3)). The function (g(x)) is decreasing when (g^\prime(x)<0). Solving (18 - 2x^{2}<0) (and (x\in[-3\sqrt{2},3\sqrt{2}])), we get (x\in[-3\sqrt{2},-3)\cup(3,3\sqrt{2}]).

Answer:

A. The function (g) is decreasing on the open interval(s) ([-3\sqrt{2},-3),(3,3\sqrt{2}])