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{2 - x^{2}}\na. on what open interval(s), if any, is the function increasing? select the correct choice below and, if necessary, fill in the answer box\na. the function g is increasing on the open interval(s) (-1,1).\n(type your answer in interval notation. type an exact answer in simplified form. use a comma to separate answers as needed\nb. the function is never increasing.\non what open interval(s), if any, is the function decreasing? select the correct choice below and, if necessary, fill in the answer box(es\na. the function g is decreasing on the open interval(s) (-\\sqrt{2},-1),(1,\\sqrt{2}).\n(type your answer in interval notation. type an exact answer in simplified form. use a comma to separate answers as needed\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 answer box(es) to complete y\n(type an exact answer in simplified form.)\na. the function has a local maximum value at three values of x. in increasing order of x - value, the maximum values are g()=\nb. the function has a local maximum value at two values of x. in increasing order of x - value, the maximum values are g(-\\sqrt{2})

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{2 - x^{2}}\na. on what open interval(s), if any, is the function increasing? select the correct choice below and, if necessary, fill in the answer box\na. the function g is increasing on the open interval(s) (-1,1).\n(type your answer in interval notation. type an exact answer in simplified form. use a comma to separate answers as needed\nb. the function is never increasing.\non what open interval(s), if any, is the function decreasing? select the correct choice below and, if necessary, fill in the answer box(es\na. the function g is decreasing on the open interval(s) (-\\sqrt{2},-1),(1,\\sqrt{2}).\n(type your answer in interval notation. type an exact answer in simplified form. use a comma to separate answers as needed\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 answer box(es) to complete y\n(type an exact answer in simplified form.)\na. the function has a local maximum value at three values of x. in increasing order of x - value, the maximum values are g()=\nb. the function has a local maximum value at two values of x. in increasing order of x - value, the maximum values are g(-\\sqrt{2})

Answer

Explanation:

Step1: Find the derivative of (g(x))

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x) and (v=\sqrt{2 - x^{2}}=(2 - x^{2})^{\frac{1}{2}}). (u^\prime=1), (v^\prime=\frac{1}{2}(2 - x^{2})^{-\frac{1}{2}}\times(- 2x)=\frac{-x}{\sqrt{2 - x^{2}}}) (g^\prime(x)=\sqrt{2 - x^{2}}+x\times\frac{-x}{\sqrt{2 - x^{2}}}=\frac{2 - x^{2}-x^{2}}{\sqrt{2 - x^{2}}}=\frac{2 - 2x^{2}}{\sqrt{2 - x^{2}}})

Step2: Find critical points

Set (g^\prime(x) = 0), then (\frac{2 - 2x^{2}}{\sqrt{2 - x^{2}}}=0). Since the denominator (\sqrt{2 - x^{2}}>0) for (x\in(-\sqrt{2},\sqrt{2})), we solve (2 - 2x^{2}=0), (x^{2}=1), (x=\pm1)

Step3: Determine increasing and decreasing intervals

  • Test the interval ((-\sqrt{2},-1)): Let (x=-1.5) (where (-\sqrt{2}\approx - 1.414)), (g^\prime(-1.5)=\frac{2-2\times(-1.5)^{2}}{\sqrt{2-(-1.5)^{2}}}=\frac{2 - 4.5}{\sqrt{2 - 2.25}}<0)
  • Test the interval ((-1,1)): Let (x = 0), (g^\prime(0)=\frac{2-2\times0^{2}}{\sqrt{2-0^{2}}}=\sqrt{2}>0)
  • Test the interval ((1,\sqrt{2})): Let (x = 1.5) (where (\sqrt{2}\approx1.414)), (g^\prime(1.5)=\frac{2-2\times(1.5)^{2}}{\sqrt{2-(1.5)^{2}}}=\frac{2 - 4.5}{\sqrt{2 - 2.25}}<0)

So (g(x)) is increasing on ((-1,1)) and decreasing on ((-\sqrt{2},-1)) and ((1,\sqrt{2}))

Step4: Find local extreme values

  • Since (g(x)) changes from decreasing to increasing at (x=-1), (g(-1)=-1\times\sqrt{2 - (-1)^{2}}=-1)
  • Since (g(x)) changes from increasing to decreasing at (x = 1), (g(1)=1\times\sqrt{2 - 1^{2}}=1)

Answer:

a. The function (g) is increasing on the open interval ((-1,1)) and decreasing on the open intervals ((-\sqrt{2},-1)) and ((1,\sqrt{2})) b. The function has local minimum value (g(-1)=-1) and local maximum value (g(1) = 1)