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√(8 - x²)\n\nb. the function is never decreasing.\nb. find each local maximum, if there are any. select the correct choice below and, if necessary, fill in the\nanswer 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()=\nb. the function has a local maximum value at two values of x. in increasing order of x - value, the\nmaximum values are g()= and g()=\nc. the function has a local maximum value at three values of x. in increasing order of x - value, the\nmaximum values are g()=, 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√(8 - x²)\n\nb. the function is never decreasing.\nb. find each local maximum, if there are any. select the correct choice below and, if necessary, fill in the\nanswer 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()=\nb. the function has a local maximum value at two values of x. in increasing order of x - value, the\nmaximum values are g()= and g()=\nc. the function has a local maximum value at three values of x. in increasing order of x - value, the\nmaximum values are g()=, 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{8 - x^{2}}), the expression under the square - root must be non - negative. So, (8-x^{2}\geqslant0), which gives (-2\sqrt{2}\leqslant x\leqslant2\sqrt{2}).

Step2: Differentiate the function using the product rule

The product rule states that if (y = u\cdot v), then (y^\prime=u^\prime v+uv^\prime). Let (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}}\cdot(-2x)=\frac{-x}{\sqrt{8 - x^{2}}}). Then (g^\prime(x)=\sqrt{8 - x^{2}}+x\cdot\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), so (\frac{8 - 2x^{2}}{\sqrt{8 - x^{2}}}=0). Since the denominator (\sqrt{8 - x^{2}}\neq0) (because if (\sqrt{8 - x^{2}} = 0), (x=\pm2\sqrt{2}) and the numerator (8 - 2x^{2}\neq0) at (x = \pm2\sqrt{2})), we solve (8 - 2x^{2}=0). (2x^{2}=8), (x^{2}=4), (x=\pm2).

Step4: Determine the intervals of increase and decrease

We use test points in the intervals ((-2\sqrt{2},-2)), ((-2,2)) and ((2,2\sqrt{2})).

  • For (x=-3) (in ((-2\sqrt{2},-2)), (g^\prime(-3)=\frac{8-2\times9}{\sqrt{8 - 9}}) (not in the domain). Let's take (x=-2.5) (not valid as (x=-2.5\notin[-2\sqrt{2},2\sqrt{2}])). Let's use the sign of (g^\prime(x)) based on the formula (g^\prime(x)=\frac{8 - 2x^{2}}{\sqrt{8 - x^{2}}}). For (x=-2) (critical point), and for (x\in(-2\sqrt{2},-2)), pick (x=- \sqrt{3}), (g^\prime(-\sqrt{3})=\frac{8-2\times3}{\sqrt{8 - 3}}=\frac{2}{\sqrt{5}}>0). For (x\in(-2,2)), pick (x = 0), (g^\prime(0)=\frac{8-0}{\sqrt{8}}>0). For (x\in(2,2\sqrt{2})), pick (x=\sqrt{6}), (g^\prime(\sqrt{6})=\frac{8 - 2\times6}{\sqrt{8 - 6}}=\frac{-4}{\sqrt{2}}<0). The function (g(x)) is increasing on ((-2\sqrt{2},2)) and decreasing on ((2,2\sqrt{2})).

Step5: Find the local and absolute extreme values

  • Local maxima: We use the first - derivative test. Since the function changes from increasing to decreasing at (x = 2). (g(2)=2\sqrt{8 - 4}=4).
  • Absolute maxima and minima: We also evaluate the function at the endpoints (x=-2\sqrt{2}) and (x = 2\sqrt{2}). (g(-2\sqrt{2})=-2\sqrt{2}\times0 = 0), (g(2\sqrt{2})=2\sqrt{2}\times0 = 0).

Answer:

a. The function (g(x)) is increasing on the interval ((-2\sqrt{2},2)) and decreasing on the interval ((2,2\sqrt{2})). b. The function has a local maximum value. The maximum value is (g(2)=4). So the answer for part b is A. The function has a local maximum value at one value of (x). The maximum value is (g(2)=4).