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\\sqrt{8 - x^{2}}\n\nfind 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.)\n\\( \\bigcirc \\) a. the function has a local maximum value at one value of \\( x \\). the maximum value is \\( g(\\quad)= \\)\n\\( \\bigcirc \\) 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 \\)\n\\( \\bigcirc \\) c. the function has a local maximum value at three values of \\( x \\). in increasing order of \\( x \\)-value, the maximum values are \\( g(\\quad)= \\), \\( g(\\quad)= \\), and \\( g(\\quad)= \\).\n\\( \\bigcirc \\) d. there are no local maxima.\nfind each local minimum, if there are any. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

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\\sqrt{8 - x^{2}}\n\nfind 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.)\n\\( \\bigcirc \\) a. the function has a local maximum value at one value of \\( x \\). the maximum value is \\( g(\\quad)= \\)\n\\( \\bigcirc \\) 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 \\)\n\\( \\bigcirc \\) c. the function has a local maximum value at three values of \\( x \\). in increasing order of \\( x \\)-value, the maximum values are \\( g(\\quad)= \\), \\( g(\\quad)= \\), and \\( g(\\quad)= \\).\n\\( \\bigcirc \\) d. there are no local maxima.\nfind each local minimum, if there are any. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

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}\geq0), which can be factored as ((2\sqrt{2}+x)(2\sqrt{2}-x)\geq0). The solutions of the inequality are (x\in[- 2\sqrt{2},2\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}}}). Then (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}}\gt0) for (x\in(-2\sqrt{2},2\sqrt{2})), we solve (8 - 2x^{2}=0). (2x^{2}=8), (x^{2} = 4), (x=\pm2). The critical points are (x=-2) and (x = 2).

Step4: Determine the intervals of increase and decrease

  • Test the interval ((-2\sqrt{2},-2)): Let (x=-3) (but (x=-3\notin[-2\sqrt{2},2\sqrt{2}])), let's take (x=-2.5) (not valid). Instead, use the fact that 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}}\gt0).
  • Test the interval ((-2,2)): Let (x = 0), (g^\prime(0)=\frac{8-0}{\sqrt{8}}=\sqrt{8}\gt0) (wrong, recalculate). Let's use the formula (g^\prime(x)=\frac{8 - 2x^{2}}{\sqrt{8 - x^{2}}}). For (x\in(-2,2)), (g^\prime(x)=\frac{8 - 2x^{2}}{\sqrt{8 - x^{2}}}). When (x = 0), (g^\prime(0)=\frac{8}{\sqrt{8}}=\sqrt{8}) (error). Correctly, (g^\prime(x)=\frac{8 - 2x^{2}}{\sqrt{8 - x^{2}}}). If (x=-1), (g^\prime(-1)=\frac{8 - 2}{\sqrt{8 - 1}}=\frac{6}{\sqrt{7}}\gt0) (wrong). Wait, (g^\prime(x)=\frac{8 - 2x^{2}}{\sqrt{8 - x^{2}}}). The sign of (g^\prime(x)) is determined by the numerator (y = 8 - 2x^{2}). The function (y = 8 - 2x^{2}) is a parabola opening downwards ((a=-2)) with (y = 0) at (x=\pm2). For (x\in(-2\sqrt{2},-2)), pick (x=-2.5) (invalid). Use the derivative formula: (g^\prime(x)=\frac{8 - 2x^{2}}{\sqrt{8 - x^{2}}}). The sign of (g^\prime(x)) is same as the sign of (8 - 2x^{2}). If (x\in(-2\sqrt{2},-2)), let (x=-2.5) (not in domain). Let's use the interval method: The critical points are (x=-2) and (x = 2). For (x\in(-2\sqrt{2},-2)), take a test point (x=-3) (invalid). Let's use the fact that (y = 8 - 2x^{2}) is positive when (|x|\lt2) and negative when (|x|\gt2) (within the domain (x\in[-2\sqrt{2},2\sqrt{2}])). (g(x)) is increasing on ((-2\sqrt{2},-2)) and ((-2,2)) (error). Wait, (g^\prime(x)=\frac{8 - 2x^{2}}{\sqrt{8 - x^{2}}}). (g^\prime(x)\gt0) when (8 - 2x^{2}\gt0) (since (\sqrt{8 - x^{2}}\gt0) for (x\in(-2\sqrt{2},2\sqrt{2}))), (x^{2}\lt4), (x\in(-2,2)). (g^\prime(x)\lt0) when (8 - 2x^{2}\lt0), (x^{2}\gt4), (x\in(-2\sqrt{2},-2)\cup(2,2\sqrt{2})).

Step5: Find the local and absolute extreme values

  • Local maxima and minima: Since (g(x)) changes from increasing to decreasing at (x = 2) (because (g^\prime(x)) changes from positive to negative at (x = 2)), (g(2)=2\sqrt{8 - 4}=4). Since (g(x)) changes from decreasing to increasing at (x=-2) (because (g^\prime(x)) changes from negative to positive at (x=-2)), (g(-2)=-2\sqrt{8 - 4}=-4).
  • Absolute maxima and minima: We also need to check the endpoints. (g(-2\sqrt{2})=-2\sqrt{2}\times0 = 0) and (g(2\sqrt{2})=2\sqrt{2}\times0 = 0).

Answer:

a. The function (g(x)) is increasing on the interval ((-2,2)) and decreasing on the intervals ((-2\sqrt{2},-2)) and ((2,2\sqrt{2})). b. The local maximum value is (g(2)=4), the local minimum value is (g(-2)=-4). The absolute maximum value is (g(2) = 4) and the absolute minimum value is (g(-2)=-4).