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{50 - x^{2}}\na. the function g is increasing on the open interval(s) (-5,5)\n(type your answer in interval notation. type an exact answer in simplified form. use a comma to\nseparate 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,\nif necessary, fill in the answer box(es) to complete your choice.\na. the function g is decreasing 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.)\nb. the function is never decreasing.

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{50 - x^{2}}\na. the function g is increasing on the open interval(s) (-5,5)\n(type your answer in interval notation. type an exact answer in simplified form. use a comma to\nseparate 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,\nif necessary, fill in the answer box(es) to complete your choice.\na. the function g is decreasing 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.)\nb. the function is never decreasing.

Answer

Explanation:

Step1: Find the domain of the function

For the function (g(x) = x\sqrt{50 - x^{2}}), the expression under the square - root must be non - negative. So, (50-x^{2}\geq0), which can be written as (x^{2}\leq50) or (-5\sqrt{2}\leq x\leq5\sqrt{2}).

Step2: Find the derivative of the function

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x) and (v=\sqrt{50 - x^{2}}=(50 - x^{2})^{\frac{1}{2}}). (u^\prime=1) and (v^\prime=\frac{1}{2}(50 - x^{2})^{-\frac{1}{2}}\times(- 2x)=\frac{-x}{\sqrt{50 - x^{2}}}) (g^\prime(x)=\sqrt{50 - x^{2}}+x\times\frac{-x}{\sqrt{50 - x^{2}}}=\frac{50 - x^{2}-x^{2}}{\sqrt{50 - x^{2}}}=\frac{50 - 2x^{2}}{\sqrt{50 - x^{2}}})

Step3: Find the critical points

Set (g^\prime(x)=0), then (50 - 2x^{2}=0) (since the denominator (\sqrt{50 - x^{2}}\neq0) for the domain (-5\sqrt{2}<x<5\sqrt{2})). (2x^{2}=50), (x^{2} = 25), (x=\pm5)

Step4: Test the intervals

  • For the interval ((-5\sqrt{2},-5)): Let (x=-6) (where (-6\in(-5\sqrt{2},-5)) and (-5\sqrt{2}\approx - 7.07)). (g^\prime(-6)=\frac{50-2\times(-6)^{2}}{\sqrt{50 - (-6)^{2}}}=\frac{50 - 72}{\sqrt{50 - 36}}=\frac{-22}{\sqrt{14}}<0)
  • For the interval ((-5,5)): Let (x = 0). (g^\prime(0)=\frac{50-2\times0^{2}}{\sqrt{50 - 0^{2}}}=\frac{50}{\sqrt{50}}=\sqrt{50}>0)
  • For the interval ((5,5\sqrt{2})): Let (x = 6) (where (6\in(5,5\sqrt{2})) and (5\sqrt{2}\approx7.07)). (g^\prime(6)=\frac{50-2\times6^{2}}{\sqrt{50 - 6^{2}}}=\frac{50 - 72}{\sqrt{50 - 36}}=\frac{-22}{\sqrt{14}}<0)

Answer:

The function (g) is decreasing on the open intervals ((-5\sqrt{2},-5)) and ((5,5\sqrt{2}))