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}}
if the function has extreme values, which of the extreme values, if any, are absolute? 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. use a comma to separate answers as needed.)
○ a. the function has no absolute maximum, but there is an absolute minimum value of at x=
○ b. the function has an absolute maximum value of at x= and an absolute minimum of at
x=
○ c. the function has an absolute maximum value of at x=, but no absolute minimum.
○ d. there are local extreme values but there are no absolute extreme values.
○ e. there are no local or absolute extreme values.

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 \(v^\prime=\frac{1}{2}(2 - x^{2})^{-\frac{1}{2}}(-2x)=\frac{-x}{\sqrt{2 - x^{2}}}\)
\(g^\prime(x)=\sqrt{2 - x^{2}}+x\cdot\frac{-x}{\sqrt{2 - x^{2}}}=\frac{2 - x^{2}-x^{2}}{\sqrt{2 - x^{2}}}=\frac{2 - 2x^{2}}{\sqrt{2 - x^{2}}}=\frac{2(1 - x^{2})}{\sqrt{2 - x^{2}}}\)

Step3: Find the critical points

Set \(g^\prime(x) = 0\), then \(2(1 - x^{2})=0\) (since \(\sqrt{2 - x^{2}}>0\) for \(-\sqrt{2}\(1 - x^{2}=0\), so \(x=- 1\) or \(x = 1\)

Step4: Determine the intervals of increase and decrease

• Interval \((-\sqrt{2},-1)\):

Pick a test point \(x=-1.5\) (not in the domain, so pick \(x = - \frac{3}{2}\) is wrong. Let's pick \(x=-1.2\)). \(g^\prime(-1.2)=\frac{2(1 - (-1.2)^{2})}{\sqrt{2-(-1.2)^{2}}}=\frac{2(1 - 1.44)}{\sqrt{2 - 1.44}}=\frac{2(-0.44)}{\sqrt{0.56}}<0\), so the function is decreasing on \((-\sqrt{2},-1)\)

• Interval \((-1,1)\):

Pick \(x = 0\), \(g^\prime(0)=\frac{2(1-0)}{\sqrt{2-0}}=\sqrt{2}>0\), so the function is increasing on \((-1,1)\)

• Interval \((1,\sqrt{2})\):

Pick \(x = 1.2\), \(g^\prime(1.2)=\frac{2(1-(1.2)^{2})}{\sqrt{2-(1.2)^{2}}}=\frac{2(1 - 1.44)}{\sqrt{2 - 1.44}}<0\), so the function is decreasing on \((1,\sqrt{2})\)

Step5: Find the extreme values

• Local and absolute values:

\(g(-\sqrt{2})=-\sqrt{2}\times0 = 0\), \(g(-1)=-1\times\sqrt{2 - 1}=-1\), \(g(1)=1\times\sqrt{2 - 1}=1\), \(g(\sqrt{2})=\sqrt{2}\times0 = 0\)

Answer:

B. The function has an absolute maximum value of \(1\) at \(x = 1\) and an absolute minimum of \(-1\) at \(x=-1\)