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

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 \(x^{2}\leq2\) or \(x\in[-\sqrt{2},\sqrt{2}]\).

Step2: Differentiate 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 critical points

Set \(g^\prime(x) = 0\), then \(1 - x^{2}=0\) (since the denominator \(\sqrt{2 - x^{2}}>0\) for \(x\in(-\sqrt{2},\sqrt{2})\)).
Solving \(1 - x^{2}=0\) gives \(x=-1\) and \(x = 1\).

Step4: Determine the sign of \(g^\prime(x)\)

• For \(x\in(-\sqrt{2},-1)\), let \(x=-1.5\) (where \(-1.5\in(-\sqrt{2},-1)\) and \(\sqrt{2}\approx1.414\)). Then \(g^\prime(-1.5)=\frac{2(1 - (-1.5)^{2})}{\sqrt{2-(-1.5)^{2}}}=\frac{2(1 - 2.25)}{\sqrt{2 - 2.25}}<0\) (the denominator is a non - real number in the real - valued function context, but if we consider the domain \(x\in[-\sqrt{2},\sqrt{2}]\) and test the intervals within the domain using test points \(x=-1.5\) is not in the domain. Let's use \(x=-1.2\) (where \(x\in(-\sqrt{2},-1)\)). \(g^\prime(-1.2)=\frac{2(1-( - 1.2)^{2})}{\sqrt{2-( - 1.2)^{2}}}=\frac{2(1 - 1.44)}{\sqrt{2 - 1.44}}=\frac{2\times(-0.44)}{\sqrt{0.56}}<0\).
• For \(x\in(-1,1)\), let \(x = 0\). Then \(g^\prime(0)=\frac{2(1-0^{2})}{\sqrt{2-0^{2}}}=\frac{2}{\sqrt{2}}>0\).
• For \(x\in(1,\sqrt{2})\), let \(x = 1.2\). Then \(g^\prime(1.2)=\frac{2(1-(1.2)^{2})}{\sqrt{2-(1.2)^{2}}}=\frac{2(1 - 1.44)}{\sqrt{2 - 1.44}}=\frac{2\times(-0.44)}{\sqrt{0.56}}<0\).

Answer:

The function \(g\) is decreasing on the open interval(s) \((-\sqrt{2},-1),(1,\sqrt{2})\)