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}}\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(-2\\sqrt{2}) = 0 and g(2)=4\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.\nfind each local minimum, if there are any. select the correct choice below and, if necessary, fill in the

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}}\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(-2\\sqrt{2}) = 0 and g(2)=4\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.\nfind each local minimum, if there are any. select the correct choice below and, if necessary, fill in the

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

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

Step4: Determine the intervals of increase and decrease

  • Test the interval ((-2\sqrt{2},-2)): Let (x=-3) (but (x=-3) is not in the domain. Let's take (x = - 2.5) (not valid). Let's use the derivative formula. Pick a test point (x=-2.5) (invalid, so we use the sign of the derivative. For (x=-2.5) (not in domain, we can also analyze the derivative (g^\prime(x)=\frac{8 - 2x^{2}}{\sqrt{8 - x^{2}}}). For (x\in(-2\sqrt{2},-2)), let (x=-2.5) (invalid, we can also note that if we consider the derivative (g^\prime(x)), when (x\in(-2\sqrt{2},-2)), (8 - 2x^{2}<0), so (g^\prime(x)<0). The function is decreasing on ((-2\sqrt{2},-2))
  • Test the interval ((-2,2)): Let (x = 0), then (g^\prime(0)=\frac{8-0}{\sqrt{8 - 0}}=\sqrt{8}>0). The function is increasing on ((-2,2))
  • Test the interval ((2,2\sqrt{2})): Let (x = 2.5) (invalid). Using the derivative (g^\prime(x)), when (x\in(2,2\sqrt{2})), (8 - 2x^{2}<0), so (g^\prime(x)<0). The function is decreasing on ((2,2\sqrt{2}))

Step5: Find the local and absolute extreme values

  • Local minima: Since the function changes from decreasing ((-2\sqrt{2},-2)) to increasing ((-2,2)), (x=-2) is a local minimum. (g(-2)=-2\sqrt{8 - 4}=-4)
  • Local maxima: Since the function changes from increasing ((-2,2)) to decreasing ((2,2\sqrt{2})), (x = 2) is a local maximum. (g(2)=2\sqrt{8 - 4}=4)

Also, (g(-2\sqrt{2})=g(2\sqrt{2})=0)

Answer:

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

b. Local minimum: (g(-2)=-4) at (x = - 2); Local maximum: (g(2)=4) at (x = 2)