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{8 - x^{2}}
d. there are no local maxima.
find each local minimum, if there are any. 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.)
a. the function has a local minimum valuę at one value of x. the minimum value is g()=
b. the function has a local minimum value at three values of x. in increasing order of x - value, the
minimum values are g()=, g()=, and g()=
c. the function has a local minimum value at two values of x. in increasing order of x - value, the
minimum values are g()= and g()=
d. there are no local minima.

Explanation:

Step1: Find the domain of the function

For the function \(g(x)=x\sqrt{8 - x^{2}}\), the expression under the square - root must be non - negative. So, \(8-x^{2}\geq0\), which gives \(-2\sqrt{2}\leq x\leq2\sqrt{2}\).

Step2: Find the derivative of the function

Using the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x\) and \(v=\sqrt{8 - x^{2}}=(8 - x^{2})^{\frac{1}{2}}\).
\(u^\prime=1\) and \(v^\prime=\frac{1}{2}(8 - x^{2})^{-\frac{1}{2}}\times(-2x)=\frac{-x}{\sqrt{8 - x^{2}}}\)
\(g^\prime(x)=\sqrt{8 - x^{2}}+x\times\frac{-x}{\sqrt{8 - x^{2}}}=\frac{8 - x^{2}-x^{2}}{\sqrt{8 - x^{2}}}=\frac{8 - 2x^{2}}{\sqrt{8 - x^{2}}}\)

Step3: Find the critical points

Set \(g^\prime(x) = 0\), then \(\frac{8 - 2x^{2}}{\sqrt{8 - x^{2}}}=0\). Since the denominator \(\sqrt{8 - x^{2}}
eq0\) (because if \(\sqrt{8 - x^{2}} = 0\), \(x=\pm2\sqrt{2}\) and the numerator \(8 - 2x^{2}
eq0\) at \(x = \pm2\sqrt{2}\)), we solve \(8 - 2x^{2}=0\).
\(2x^{2}=8\), \(x^{2}=4\), \(x=\pm2\)

Step4: Analyze the sign of the derivative

• When \(-2\sqrt{2}
• For the interval \((-2\sqrt{2},-2)\), pick \(x=-2.5\) (invalid, pick \(x = - 2.1\) (not in the domain). Let's use the fact that \(g^\prime(x)=\frac{8 - 2x^{2}}{\sqrt{8 - x^{2}}}\). If \(x=-2.1\) (not in domain, better: when \(x=-2\), \(g^\prime(x) = 0\). For \(x\in(-2\sqrt{2},-2)\), let \(x=-2.5\) (invalid, use \(x=-2.1\) (not in domain). Let's use the substitution \(y = x^{2}\). The function \(y = 8 - 2x^{2}\) is a parabola opening downwards.
• For \(x\in(-2\sqrt{2},-2)\), \(g^\prime(x)<0\) (since \(8 - 2x^{2}<0\) when \(x^{2}>4\) and \(x\in(-2\sqrt{2},-2)\)).
• For \(x\in(-2,2)\), \(g^\prime(x)>0\) (since \(8 - 2x^{2}>0\) when \(x^{2}<4\)).
• For \(x\in(2,2\sqrt{2})\), \(g^\prime(x)<0\) (since \(8 - 2x^{2}<0\) when \(x^{2}>4\)).

Step5: Find local extrema

• Since the function \(g(x)\) changes from decreasing \((-2\sqrt{2},-2)\) to increasing \((-2,2)\), \(x=-2\) is a local minimum.

\(g(-2)=-2\sqrt{8 - 4}=-2\times2=-4\)

• Since the function \(g(x)\) changes from increasing \((-2,2)\) to decreasing \((2,2\sqrt{2})\), \(x = 2\) is a local maximum. \(g(2)=2\sqrt{8 - 4}=4\)

Answer:

A. The function has a local minimum value at one value of \(x\). The minimum value is \(g(-2)=-4\)