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, \(2-x^{2}\geqslant0\), which gives \(-\sqrt{2}\leqslant x\leqslant\sqrt{2}\).

Step2: Differentiate the function using the product rule

The product rule states that if \(y = u\cdot v\), then \(y^\prime=u^\prime v + uv^\prime\). Let \(u = x\) and \(v=\sqrt{2 - x^{2}}=(2 - x^{2})^{\frac{1}{2}}\).
\(u^\prime = 1\) and \(v^\prime=\frac{1}{2}(2 - x^{2})^{-\frac{1}{2}}\cdot(- 2x)=\frac{-x}{\sqrt{2 - x^{2}}}\)
\(g^\prime(x)=\sqrt{2 - x^{2}}+x\cdot\frac{-x}{\sqrt{2 - x^{2}}}=\frac{2 - x^{2}-x^{2}}{\sqrt{2 - x^{2}}}=\frac{2 - 2x^{2}}{\sqrt{2 - x^{2}}}\)

Step3: Find the critical points

Set \(g^\prime(x)=0\), then \(\frac{2 - 2x^{2}}{\sqrt{2 - x^{2}}}=0\). Since the denominator \(\sqrt{2 - x^{2}}>0\) for \(x\in(-\sqrt{2},\sqrt{2})\), we solve \(2 - 2x^{2}=0\).
\(2x^{2}=2\), \(x^{2} = 1\), \(x=\pm1\)

Step4: Test the intervals

We test the intervals \((-\sqrt{2},- 1)\), \((-1,1)\) and \((1,\sqrt{2})\)

• For \(x\in(-\sqrt{2},-1)\), let \(x=-1.5\) (where \(-\sqrt{2}\approx - 1.414\)). \(g^\prime(-1.5)=\frac{2-2\times(-1.5)^{2}}{\sqrt{2-(-1.5)^{2}}}=\frac{2 - 4.5}{\sqrt{2 - 2.25}}<0\)
• For \(x\in(-1,1)\), let \(x = 0\). \(g^\prime(0)=\frac{2-2\times0^{2}}{\sqrt{2-0^{2}}}=\frac{2}{\sqrt{2}}>0\)
• For \(x\in(1,\sqrt{2})\), let \(x = 1.5\). \(g^\prime(1.5)=\frac{2-2\times(1.5)^{2}}{\sqrt{2-(1.5)^{2}}}=\frac{2 - 4.5}{\sqrt{2 - 2.25}}<0\)

Answer:

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