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{2 - x^{2}}
a. on what open interval(s), if any, is the function increasing? select the correct choice below and,
if necessary, fill in the answer box(es) to complete your choice.
○ a. the function g is increasing 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 increasing.

Explanation:

Step1: Find the domain of the function

For the function \(g(x)=x\sqrt{2 - x^{2}}\), the expression under the square - root must be non - negative. So, \(2-x^{2}\geq0\), which gives \(-\sqrt{2}\leq x\leq\sqrt{2}\).

Step2: Find the derivative of the function

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

Step3: Find where \(g^\prime(x)>0\)

Set \(g^\prime(x)>0\), \(\frac{2(1 - x^{2})}{\sqrt{2 - x^{2}}}>0\).
Since the denominator \(\sqrt{2 - x^{2}}>0\) for \(-\sqrt{2}0\).
Solving \(1 - x^{2}>0\) (i.e., \((1 - x)(1 + x)>0\)), we get \(- 1

Answer:

A. The function \(g\) is increasing on the open interval(s) \((-1,1)\)