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{18 - x^{2}}\nd. find each local maximum, if there are any. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.\n(type an exact answer in simplified form.)\na. the function has a local maximum value at one value of x. the maximum value is 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{18}) = 0 and g()=\nc. the function has a local maximum value at three values of x. in increasing order of x - value, the maximum values are g() =, g() =, and g()=\nd. there are no local maxima.

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{18 - x^{2}}\nd. find each local maximum, if there are any. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.\n(type an exact answer in simplified form.)\na. the function has a local maximum value at one value of x. the maximum value is 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{18}) = 0 and g()=\nc. the function has a local maximum value at three values of x. in increasing order of x - value, the maximum values are g() =, g() =, and g()=\nd. there are no local maxima.

Answer

Explanation:

Step1: Find the domain and derivative

The domain of (g(x)=x\sqrt{18 - x^{2}}) is (-3\sqrt{2}\leq x\leq3\sqrt{2}) since (18 - x^{2}\geq0). Using the product rule ((uv)^\prime = u^\prime v+uv^\prime) where (u = x) and (v=\sqrt{18 - x^{2}}=(18 - x^{2})^{\frac{1}{2}}). (u^\prime=1) and (v^\prime=\frac{-x}{\sqrt{18 - x^{2}}}) (g^\prime(x)=\sqrt{18 - x^{2}}+x\cdot\frac{-x}{\sqrt{18 - x^{2}}}=\frac{18 - 2x^{2}}{\sqrt{18 - x^{2}}})

Step2: Find critical points

Set (g^\prime(x) = 0), so (\frac{18 - 2x^{2}}{\sqrt{18 - x^{2}}}=0) (since the denominator (\sqrt{18 - x^{2}}>0) for (-3\sqrt{2}<x<3\sqrt{2})). (18 - 2x^{2}=0) gives (x^{2} = 9), (x=\pm3)

Step3: Test intervals for increasing/decreasing

  • For (-3\sqrt{2}<x<-3), let (x=-4) (not in domain, actually test (x = - 3.5) is wrong, better use test points in domain: Take a test point (x=-2) in ((-3\sqrt{2},-3)), (g^\prime(-2)=\frac{18-2\times(-2)^{2}}{\sqrt{18-(-2)^{2}}}=\frac{18 - 8}{\sqrt{14}}>0) (wrong, correct: Take test points: For the interval ((-3\sqrt{2},-3)), pick (x=- 4) is wrong, pick (x=-3.5) is wrong, actually for (y = g^\prime(x)=\frac{18 - 2x^{2}}{\sqrt{18 - x^{2}}}), when (x\in(-3\sqrt{2},-3)), (18-2x^{2}<0) (since (x^{2}>9)), (g^\prime(x)<0) For (x\in(-3,3)), (18 - 2x^{2}>0), (g^\prime(x)>0) For (x\in(3,3\sqrt{2})), (18-2x^{2}<0), (g^\prime(x)<0)

Step4: Find local maxima

Since (g(x)) changes from decreasing ((-3\sqrt{2},-3)) to increasing ((-3,3)) and then to decreasing ((3,3\sqrt{2})) (g(-3)=-3\sqrt{18 - 9}=-9) (g(3)=3\sqrt{18 - 9}=9)

Answer:

A. The function has a local maximum value at one value of (x). The maximum value is (g(3)=9)