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}}
answer box(es) to complete your choice.
(type an exact answer in simplified form.)
a. the function has a local maximum value at one value of x. the maximum value is g( )=
b. the function has a local maximum value at two values of x. in increasing order of x - value, the
maximum values are g(-2\sqrt{2}) = 0 and g(2)=4
c. the function has a local maximum value at three values of x. in increasing order of x - value, the
maximum values are g( )=,g( )=, and g( )=
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

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 \(\sqrt{8 - x^{2}}>0\) for \(-2\sqrt{2}

\(2x^{2}=8\), \(x^{2} = 4\), \(x=\pm2\)

Step4: Determine the intervals of increase and decrease

• Test the interval \((-2\sqrt{2},-2)\):

Let \(x=-3\) (but \(x=-3\) is not in the domain. Let's take \(x = - 2.5\) (not valid). Let's use the derivative formula.
Pick a test point \(x=-2.5\) (invalid, so we use the sign of the derivative. For \(x=-2.5\) (not in domain, we can also analyze the derivative \(g^\prime(x)=\frac{8 - 2x^{2}}{\sqrt{8 - x^{2}}}\).
For \(x\in(-2\sqrt{2},-2)\), let \(x=-2.5\) (invalid, we can also note that if we consider the derivative \(g^\prime(x)\), when \(x\in(-2\sqrt{2},-2)\), \(8 - 2x^{2}<0\), so \(g^\prime(x)<0\). The function is decreasing on \((-2\sqrt{2},-2)\)

• Test the interval \((-2,2)\):

Let \(x = 0\), then \(g^\prime(0)=\frac{8-0}{\sqrt{8 - 0}}=\sqrt{8}>0\). The function is increasing on \((-2,2)\)

• Test the interval \((2,2\sqrt{2})\):

Let \(x = 2.5\) (invalid). Using the derivative \(g^\prime(x)\), when \(x\in(2,2\sqrt{2})\), \(8 - 2x^{2}<0\), so \(g^\prime(x)<0\). The function is decreasing on \((2,2\sqrt{2})\)

Step5: Find the local and absolute extreme values

• Local minima:

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

• Local maxima:

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

Also, \(g(-2\sqrt{2})=g(2\sqrt{2})=0\)

Answer:

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

b. Local minimum: \(g(-2)=-4\) at \(x = - 2\); Local maximum: \(g(2)=4\) at \(x = 2\)