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 } }
separate answers as needed.)
b. the function is never increasing.
on what open interval(s), if any, is the function decrelasing? 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 and derivative

The domain of \(g(x)=x\sqrt{2 - x^{2}}\) is \([-\sqrt{2},\sqrt{2}]\) since \(2-x^{2}\geq0\).
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 \(v^\prime=\frac{-2x}{2\sqrt{2 - x^{2}}}=\frac{-x}{\sqrt{2 - x^{2}}}\).
Then \(g^\prime(x)=\sqrt{2 - x^{2}}+x\cdot\frac{-x}{\sqrt{2 - x^{2}}}=\frac{2 - x^{2}-x^{2}}{\sqrt{2 - x^{2}}}=\frac{2 - 2x^{2}}{\sqrt{2 - x^{2}}}=\frac{2(1 - x^{2})}{\sqrt{2 - x^{2}}}\).

Step2: Find 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 \(x\in(-\sqrt{2},\sqrt{2})\), we solve \(1 - x^{2}=0\), so \(x=- 1\) or \(x = 1\).

Step3: Determine increasing and decreasing intervals

Test intervals:

• For \(x\in(-\sqrt{2},-1)\), let \(x=-1.5\) (not in domain, adjust to \(x = - 1.2\)), \(g^\prime(-1.2)=\frac{2(1-( - 1.2)^{2})}{\sqrt{2-( - 1.2)^{2}}}=\frac{2(1 - 1.44)}{\sqrt{2 - 1.44}}=\frac{2(-0.44)}{\sqrt{0.56}}<0\).
• For \(x\in(-1,1)\), let \(x = 0\), \(g^\prime(0)=\frac{2(1-0)}{\sqrt{2-0}}=\sqrt{2}>0\).
• For \(x\in(1,\sqrt{2})\), let \(x = 1.2\), \(g^\prime(1.2)=\frac{2(1-(1.2)^{2})}{\sqrt{2-(1.2)^{2}}}=\frac{2(1 - 1.44)}{\sqrt{2 - 1.44}}=\frac{2(-0.44)}{\sqrt{0.56}}<0\).

The function \(g(x)\) is increasing on \((-1,1)\) and decreasing on \((-\sqrt{2},-1)\cup(1,\sqrt{2})\).

Answer:

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