Question

a. find the open interval(s) on which the function is increasing and decreasing
b. identify the functions local and absolute extreme values, if any, saying where they occur
g(x)=x\sqrt{8 - x^{2}}
a. the function g is increasing on the open interval(s) (-2,2)
(type your answer in interval notation. type an exact answer in simplified form. use a comma to
separate answers as needed.)
b. the function is never increasing.
on what open interval(s), if any, is the function decreasing? select the correct choice below and,
if necessary, fill in the answer box(es) to complete your choice.
a. the function g is decreasing on the open interval(s)
(type your answer in interval notation. type an exact answer in simplified form. use a comma to
separate answers as needed.)
b. the function is never decreasing.

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 solutions of the inequality \(8 - x^{2}\geq0\) are \(x\in[- 2\sqrt{2},2\sqrt{2}]\).

Step2: Find the derivative of the function

Using the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x\) and \(v=\sqrt{8 - x^{2}}=(8 - x^{2})^{\frac{1}{2}}\).
The derivative of \(u=x\) is \(u^\prime = 1\).
Using the chain rule, the derivative of \(v=(8 - x^{2})^{\frac{1}{2}}\) is \(v^\prime=\frac{1}{2}(8 - x^{2})^{-\frac{1}{2}}\times(-2x)=\frac{-x}{\sqrt{8 - x^{2}}}\).
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 \(x=-3\) (a test point). Then \(g^\prime(-3)=\frac{8-2\times(-3)^{2}}{\sqrt{8 - (-3)^{2}}}=\frac{8 - 18}{\sqrt{-1}}\) (not in the domain). Let's use the fact that \(g^\prime(x)=\frac{8 - 2x^{2}}{\sqrt{8 - x^{2}}}\). If \(x\in(-2\sqrt{2},-2)\), say \(x=-2.5\) (approximate value in the domain), \(8-2x^{2}=8-2\times6.25=-4.5<0\), so \(g^\prime(x)<0\) on \((-2\sqrt{2},-2)\).
• For \(x\in(-2,2)\), let \(x = 0\), then \(g^\prime(0)=\frac{8-0}{\sqrt{8-0}}=\sqrt{8}>0\).
• For \(x\in(2,2\sqrt{2})\), let \(x = 2.5\) (approximate value in the domain), \(8-2x^{2}=8 - 2\times6.25=-4.5<0\), so \(g^\prime(x)<0\) on \((2,2\sqrt{2})\).

Answer:

The function \(g\) is decreasing on the open interval(s) \((-2\sqrt{2},-2),(2,2\sqrt{2})\)