a. find the open interval(s) on which the function is increasing and decreasing\nb. identify the functions…

a. find the open interval(s) on which the function is increasing and decreasing\nb. identify the functions local and absolute extreme values, if any, saying where they occur\ng(x)=x√(18 - x²)\nb. 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.\n(type an exact answer in simplified form.)\na. the function has a local maximum value at one value of x. the maximum value is g(3)=9\nb. 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()=\nc. 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()=\nd. there are no local maxima.

a. find the open interval(s) on which the function is increasing and decreasing\nb. identify the functions local and absolute extreme values, if any, saying where they occur\ng(x)=x√(18 - x²)\nb. 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.\n(type an exact answer in simplified form.)\na. the function has a local maximum value at one value of x. the maximum value is g(3)=9\nb. 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()=\nc. 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()=\nd. there are no local maxima.

Answer

Explanation:

Step1: Find the domain of the function

For the function (g(x)=x\sqrt{18 - x^{2}}), the expression under the square - root must be non - negative. So, (18-x^{2}\geq0), which gives ((x + 3\sqrt{2})(x-3\sqrt{2})\leq0). The domain is ([- 3\sqrt{2},3\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{18 - x^{2}}=(18 - x^{2})^{\frac{1}{2}}). (u^\prime=1) and (v^\prime=\frac{1}{2}(18 - x^{2})^{-\frac{1}{2}}\times(-2x)=\frac{-x}{\sqrt{18 - x^{2}}}). (g^\prime(x)=\sqrt{18 - x^{2}}+x\times\frac{-x}{\sqrt{18 - x^{2}}}=\frac{18 - x^{2}-x^{2}}{\sqrt{18 - x^{2}}}=\frac{18 - 2x^{2}}{\sqrt{18 - x^{2}}}).

Step3: Find the critical points

Set (g^\prime(x) = 0), then (18-2x^{2}=0) (since the denominator (\sqrt{18 - x^{2}}\neq0) in the interior of the domain). (2x^{2}=18), (x^{2}=9), (x=\pm3).

Step4: Determine the intervals of increase and decrease

  • For the interval ((-3\sqrt{2},-3)), let (x=-4) (a test point). (g^\prime(-4)=\frac{18-2\times16}{\sqrt{18 - 16}}=\frac{18 - 32}{\sqrt{2}}=\frac{-14}{\sqrt{2}}<0). So, (g(x)) is decreasing on ((-3\sqrt{2},-3)).
  • For the interval ((-3,3)), let (x = 0) (a test point). (g^\prime(0)=\frac{18-0}{\sqrt{18}}=\sqrt{18}>0). So, (g(x)) is increasing on ((-3,3)).
  • For the interval ((3,3\sqrt{2})), let (x = 4) (a test point). (g^\prime(4)=\frac{18-2\times16}{\sqrt{18 - 16}}=\frac{18 - 32}{\sqrt{2}}=\frac{-14}{\sqrt{2}}<0). So, (g(x)) is decreasing on ((3,3\sqrt{2})).

Step5: Find the local maxima and minima

Since the function changes from decreasing to increasing at (x=-3) and from increasing to decreasing at (x = 3). (g(-3)=-3\sqrt{18 - 9}=-9) and (g(3)=3\sqrt{18 - 9}=9).

Answer:

A. The function has a local maximum value at one value of (x). The maximum value is (g(3)=9)