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}}
b. find each local maximum, if there are any. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.
(type an exact answer in simplified form.)
○ a. the function has a local maximum value at three values of x. in increasing order of x - value, the maximum values are g(\square)=\square, g(\square)=\square, and g(\square)=\square.
○ b. the function has a local maximum value at two values of x. in increasing order of x - value, the maximum values are g(\square)=\square and g(\square)=\square.
○ c. the function has a local maximum value at one value of x. the maximum value is g(\square)=\square
○ d. there are no local maxima.

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 the critical points

Set \(g^\prime(x) = 0\), then \(\frac{2(1 - x^{2})}{\sqrt{2 - x^{2}}}=0\). Since the denominator \(\sqrt{2 - x^{2}}>0\) for \(-\sqrt{2}

Step4: Determine the intervals of increase and decrease

• For \(-\sqrt{2}
• For \(-1
• For \(1

So, the function \(g(x)\) is increasing on the interval \((-1,1)\) and decreasing on the intervals \((-\sqrt{2},-1)\) and \((1,\sqrt{2})\).

Step5: Find the local extreme values

• Evaluate \(g(x)\) at the critical points:
• \(g(-1)=-1\times\sqrt{2-(-1)^{2}}=-1\).
• \(g(1)=1\times\sqrt{2 - 1^{2}}=1\).

Answer:

a. The function \(g(x)\) is increasing on the interval \((-1,1)\) and decreasing on the intervals \((-\sqrt{2},-1)\) and \((1,\sqrt{2})\).
b. The function has a local maximum value at one value of \(x\). The maximum value is \(g(1)=1\) (Option C).