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. on what open interval(s), if any, is the function increasing? select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.\na. the function g is increasing on the open interval(s) (type your answer in interval notation. type an exact answer in simplified form. use a comma separate answers as needed.)\nb. 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{50 - 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(es) to complete your choice.\na. the function g is increasing on the open interval(s) (type your answer in interval notation. type an exact answer in simplified form. use a comma separate answers as needed.)\nb. the function is never increasing

Answer

Explanation:

Step1: Determine the domain

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 gives (x^{2}\leq50) or (x\in[-\sqrt{50},\sqrt{50}]=[-5\sqrt{2},5\sqrt{2}]).

Step2: Find the derivative

Using 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}}}). Then (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 critical points

Set (g^\prime(x) = 0), so (\frac{50 - 2x^{2}}{\sqrt{50 - x^{2}}}=0). Since the denominator (\sqrt{50 - x^{2}}>0) for (x\in(-5\sqrt{2},5\sqrt{2})), we solve (50 - 2x^{2}=0). (2x^{2}=50), (x^{2} = 25), (x=\pm5).

Step4: Test intervals

  • For the interval ((-5\sqrt{2},- 5)), let (x=-6) (but (x=-6\notin(-5\sqrt{2},-5)) since (-5\sqrt{2}\approx - 7.07)). Let's use a test point (x=-6) (invalid, instead take (x=-6) is wrong. Let's take (x=-6) is wrong. Let's take (x = - 6) (wrong approach). Let's use the sign of (g^\prime(x)) formula. The derivative (g^\prime(x)=\frac{50 - 2x^{2}}{\sqrt{50 - x^{2}}}). For (x\in(-5\sqrt{2},-5)), pick (x=-6) (invalid, correct way: consider the function (y = 50-2x^{2}), when (x\in(-5\sqrt{2},-5)), (y = 50-2x^{2}<0) (because (x^{2}>25) for (|x|>5)). For (x\in(-5,5)), pick (x = 0), then (g^\prime(0)=\frac{50-0}{\sqrt{50 - 0}}=\sqrt{50}>0). For (x\in(5,5\sqrt{2})), pick (x = 6) (invalid, consider (y = 50-2x^{2}), when (x\in(5,5\sqrt{2})), (x^{2}>25), so (y = 50 - 2x^{2}<0).

Answer:

A. The function (g) is increasing on the open interval((-5,5))