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√{50 - x²}
ob. the function is never decreasing.
b. find each local maximum, 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.)
oa. the function has a local maximum value at one value of x. the maximum value is g ( ) =
ob. 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 ( ) =
c. the function has a local maximum value at two values of x. in increasing order of x - value, the
maximum values are g ( ) = and g ( ) =
od. there are no local maxima.

Explanation:

Step1: Find the domain of the function

For the function \(g(x)=x\sqrt{50 - x^{2}}\), the expression under the square - root must be non - negative. So, \(50-x^{2}\geq0\), which gives \(-5\sqrt{2}\leq x\leq5\sqrt{2}\).

Step2: Differentiate the function using the product rule

The product rule is \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x\) and \(v=\sqrt{50 - x^{2}}=(50 - x^{2})^{\frac{1}{2}}\).
\(u^\prime=1\) and \(v^\prime=\frac{1}{2}(50 - x^{2})^{-\frac{1}{2}}\times(-2x)=\frac{-x}{\sqrt{50 - x^{2}}}\).
Then \(g^\prime(x)=\sqrt{50 - x^{2}}+x\times\frac{-x}{\sqrt{50 - x^{2}}}=\frac{50 - x^{2}-x^{2}}{\sqrt{50 - x^{2}}}=\frac{50 - 2x^{2}}{\sqrt{50 - x^{2}}}\).

Step3: Find the critical points

Set \(g^\prime(x) = 0\), so \(\frac{50 - 2x^{2}}{\sqrt{50 - x^{2}}}=0\). Since the denominator \(\sqrt{50 - x^{2}}>0\) for \(-5\sqrt{2}\(2x^{2}=50\), \(x^{2}=25\), \(x=\pm5\).

Step4: Determine the intervals of increase and decrease

We use test points in the intervals \((-5\sqrt{2},-5)\), \((-5,5)\) and \((5,5\sqrt{2})\).
For \(x=-6\) (in \((-5\sqrt{2},-5)\)), \(g^\prime(-6)=\frac{50-2\times(-6)^{2}}{\sqrt{50 - (-6)^{2}}}=\frac{50 - 72}{\sqrt{50 - 36}}=\frac{-22}{\sqrt{14}}<0\).
For \(x = 0\) (in \((-5,5)\)), \(g^\prime(0)=\frac{50-2\times0^{2}}{\sqrt{50 - 0^{2}}}=\frac{50}{\sqrt{50}}=\sqrt{50}>0\).
For \(x = 6\) (in \((5,5\sqrt{2})\)), \(g^\prime(6)=\frac{50-2\times6^{2}}{\sqrt{50 - 6^{2}}}=\frac{50 - 72}{\sqrt{50 - 36}}=\frac{-22}{\sqrt{14}}<0\).

So, the function \(g(x)\) is decreasing on \((-5\sqrt{2},-5)\) and \((5,5\sqrt{2})\), and increasing on \((-5,5)\).

Step5: Find the local extreme values

Since the function changes from decreasing to increasing at \(x=-5\) and from increasing to decreasing at \(x = 5\).
\(g(-5)=-5\sqrt{50-(-5)^{2}}=-5\sqrt{25}=-25\).
\(g(5)=5\sqrt{50 - 5^{2}}=5\sqrt{25}=25\).

Answer:

a. The function \(g(x)\) is decreasing on \((-5\sqrt{2},-5)\cup(5,5\sqrt{2})\) and increasing on \((-5,5)\).
b. The function has a local maximum value at \(x = 5\) and \(g(5)=25\), and a local minimum value at \(x=-5\) and \(g(-5)=-25\). So, for the local maximum part (if we only consider the local maximum from the options related to local maximum in the problem's b - part format), the function has a local maximum value at one value of \(x\) (if we consider the non - multiple - choice for local maximum in a general sense, but if we follow the problem's option C which is wrongly formatted in the original problem's context. In fact, using the first - derivative test):
The function has a local maximum at \(x = 5\) where \(g(5)=25\) and a local minimum at \(x=-5\) where \(g(-5)=-25\). If we only focus on local maximum for the problem's b - part (assuming the problem's options are mis - labeled and we consider the correct local maximum): The function has a local maximum value at one value of \(x\). The maximum value is \(g(5)=25\).