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√(8 - x²)
answer box(es) to complete your choice.
(type an exact answer in simplified form.)
a. the function has a local minimum value 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()=0 and g(-2)= - 4
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 can be factored as \((2\sqrt{2}+x)(2\sqrt{2}-x)\geq0\). The solution is \(- 2\sqrt{2}\leq x\leq2\sqrt{2}\).

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 \(x\in(-2\sqrt{2},2\sqrt{2})\), we solve \(8 - 2x^{2}=0\).

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

Step4: Analyze the sign of the derivative

• For \(x\in(-2\sqrt{2},-2)\), let \(x=-3\) (not in the domain, but we can test a value close to \(-2\sqrt{2}\) like \(x = - 2.5\) (not in the domain, better use test points in the domain). Let \(x=-2.8\) (invalid, use \(x=-2.1\)). \(g^\prime(-2.1)=\frac{8-2\times(-2.1)^{2}}{\sqrt{8-(-2.1)^{2}}}=\frac{8 - 2\times4.41}{\sqrt{8 - 4.41}}=\frac{8 - 8.82}{\sqrt{3.59}}<0\)
• For \(x\in(-2,2)\), let \(x = 0\), \(g^\prime(0)=\frac{8-0}{\sqrt{8}}>0\)
• For \(x\in(2,2\sqrt{2})\), let \(x = 2.1\), \(g^\prime(2.1)=\frac{8-2\times(2.1)^{2}}{\sqrt{8-(2.1)^{2}}}=\frac{8 - 8.82}{\sqrt{3.59}}<0\)

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

Step5: Find the local extrema

• Since the function changes from decreasing to increasing at \(x=-2\), \(g(-2)=-2\sqrt{8 - 4}=-4\)
• Since the function changes from increasing to decreasing at \(x = 2\), \(g(2)=2\sqrt{8 - 4}=4\)

Answer:

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