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)\nanswer box(es) to complete your choice.\n(type an exact answer in simplified form.)\na. the function has a local minimum value at one value of x. the minimum value is g()=\nb. the function has a local minimum value at three values of x. in increasing order of x - value, the minimum values are g()=, g()=, and g()=\nc. the function has a local minimum value at two values of x. in increasing order of x - value, the minimum values are g()=0 and g(-2)= - 4\nd. there are no local minima

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)\nanswer box(es) to complete your choice.\n(type an exact answer in simplified form.)\na. the function has a local minimum value at one value of x. the minimum value is g()=\nb. the function has a local minimum value at three values of x. in increasing order of x - value, the minimum values are g()=, g()=, and g()=\nc. the function has a local minimum value at two values of x. in increasing order of x - value, the minimum values are g()=0 and g(-2)= - 4\nd. there are no local minima

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 solution is (- 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 (x\in(-2\sqrt{2},2\sqrt{2})), we solve (8 - 2x^{2}=0).

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

Step4: Analyze the sign of the derivative

  • For (x\in(-2\sqrt{2},-2)), let (x=-3) (not in the domain, but we can test a value close to (-2\sqrt{2}) like (x = - 2.5) (not in the domain, better use test points in the domain). Let (x=-2.8) (invalid, use (x=-2.1)). (g^\prime(-2.1)=\frac{8-2\times(-2.1)^{2}}{\sqrt{8-(-2.1)^{2}}}=\frac{8 - 2\times4.41}{\sqrt{8 - 4.41}}=\frac{8 - 8.82}{\sqrt{3.59}}<0)
  • For (x\in(-2,2)), let (x = 0), (g^\prime(0)=\frac{8-0}{\sqrt{8}}>0)
  • For (x\in(2,2\sqrt{2})), let (x = 2.1), (g^\prime(2.1)=\frac{8-2\times(2.1)^{2}}{\sqrt{8-(2.1)^{2}}}=\frac{8 - 8.82}{\sqrt{3.59}}<0)

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

Step5: Find the local extrema

  • Since the function changes from decreasing to increasing at (x=-2), (g(-2)=-2\sqrt{8 - 4}=-4)
  • Since the function changes from increasing to decreasing at (x = 2), (g(2)=2\sqrt{8 - 4}=4)

Answer:

A. The function has a local minimum value at one value of (x). The minimum value is (g(-2)=-4)