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 { 18 - 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

oa. 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.)

ob. the function is never increasing

Explanation:

Step1: Find the domain of the function

For the function \(g(x)=x\sqrt{18 - x^{2}}\), the expression under the square - root must be non - negative. So, \(18-x^{2}\geq0\), which can be factored as \((\sqrt{18}+x)(\sqrt{18}-x)\geq0\). Solving \(18 - x^{2}=0\) gives \(x =-\sqrt{18}=-3\sqrt{2}\) and \(x = 3\sqrt{2}\). The domain of \(g(x)\) is \([-3\sqrt{2},3\sqrt{2}]\).

Step2: Find the derivative of the function

Use the product rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u = x\) and \(v=\sqrt{18 - x^{2}}=(18 - x^{2})^{\frac{1}{2}}\).
\(u^\prime=1\) and \(v^\prime=\frac{1}{2}(18 - x^{2})^{-\frac{1}{2}}\times(-2x)=\frac{-x}{\sqrt{18 - x^{2}}}\).
Then \(g^\prime(x)=\sqrt{18 - x^{2}}+x\times\frac{-x}{\sqrt{18 - x^{2}}}=\frac{18 - x^{2}-x^{2}}{\sqrt{18 - x^{2}}}=\frac{18 - 2x^{2}}{\sqrt{18 - x^{2}}}\).

Step3: Find the critical points

Set \(g^\prime(x) = 0\), so \(\frac{18 - 2x^{2}}{\sqrt{18 - x^{2}}}=0\). Since the denominator \(\sqrt{18 - x^{2}}>0\) for \(x\in(-3\sqrt{2},3\sqrt{2})\), we solve \(18 - 2x^{2}=0\).
\(2x^{2}=18\), \(x^{2}=9\), \(x=\pm3\).

Step4: Determine the sign of the derivative

We use test points in the intervals \((-3\sqrt{2},-3)\), \((-3,3)\) and \((3,3\sqrt{2})\).
For the interval \((-3\sqrt{2},-3)\), let \(x=-4\) (but \(x=-4
otin[-3\sqrt{2},3\sqrt{2}]\)), for the interval \((-3,3)\), let \(x = 0\). Then \(g^\prime(0)=\frac{18-0}{\sqrt{18-0}}=\sqrt{18}>0\).
For the interval \((3,3\sqrt{2})\), let \(x = 4\) (but \(x = 4
otin[-3\sqrt{2},3\sqrt{2}]\)).

Answer:

A. The function \(g\) is increasing on the open interval \((-3,3)\)