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,\nif necessary, fill in the answer box(es) to complete your choice.\n○ a. the function g is increasing on the open interval(s)\n(type your answer in interval notation. type an exact answer in simplified form. use a comma to\nseparate answers as needed.)\n○ b. the function is never increasing

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,\nif necessary, fill in the answer box(es) to complete your choice.\n○ a. the function g is increasing on the open interval(s)\n(type your answer in interval notation. type an exact answer in simplified form. use a comma to\nseparate answers as needed.)\n○ b. the function is never increasing

Answer

Explanation:

Step1: Find the domain of the function

For the function (g(x) = x\sqrt{2 - x^{2}}), the expression under the square - root must be non - negative. So, (2-x^{2}\geqslant0), which gives (-\sqrt{2}\leqslant x\leqslant\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{2 - x^{2}}=(2 - x^{2})^{\frac{1}{2}}). (u^\prime = 1) and (v^\prime=\frac{1}{2}(2 - x^{2})^{-\frac{1}{2}}\cdot(- 2x)=\frac{-x}{\sqrt{2 - x^{2}}}) (g^\prime(x)=\sqrt{2 - x^{2}}+x\cdot\frac{-x}{\sqrt{2 - x^{2}}}=\frac{2 - x^{2}-x^{2}}{\sqrt{2 - x^{2}}}=\frac{2 - 2x^{2}}{\sqrt{2 - x^{2}}})

Step3: Find the 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). (2x^{2}=2), (x^{2} = 1), (x=\pm1)

Step4: Test the intervals

We test the intervals ((-\sqrt{2},- 1)), ((-1,1)) and ((1,\sqrt{2}))

  • For (x\in(-\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)
  • For (x\in(-1,1)), let (x = 0). (g^\prime(0)=\frac{2-2\times0^{2}}{\sqrt{2-0^{2}}}=\frac{2}{\sqrt{2}}>0)
  • For (x\in(1,\sqrt{2})), let (x = 1.5). (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)

Answer:

A. The function (g) is increasing on the open interval ((-1,1))