identify the coordinates of any local and absolute extreme points, inflection points, and intervals of…

identify the coordinates of any local and absolute extreme points, inflection points, and intervals of concavity. graph the function.\nf(x)=\\ln\\left(30 - 2x^{2}\\right)\nidentify the absolute maximum value and where it occurs. select the correct choice and, if necessary, fill in the answer box to complete your choice.\n○ a. the absolute maximum value \\square occurs at x = \\square.\n(use a comma to separate answers as needed. type each answer only once.)\n○ b. there is no absolute maximum point.

identify the coordinates of any local and absolute extreme points, inflection points, and intervals of concavity. graph the function.\nf(x)=\\ln\\left(30 - 2x^{2}\\right)\nidentify the absolute maximum value and where it occurs. select the correct choice and, if necessary, fill in the answer box to complete your choice.\n○ a. the absolute maximum value \\square occurs at x = \\square.\n(use a comma to separate answers as needed. type each answer only once.)\n○ b. there is no absolute maximum point.

Answer

Explanation:

Step1: Find the domain

For (y = \ln(30 - 2x^{2})), the argument of the logarithm must be positive. So (30-2x^{2}>0), which gives (x^{2}<15) or (-\sqrt{15}<x<\sqrt{15}).

Step2: Find the first - derivative

Using the chain rule, if (y=\ln(u)) and (u = 30 - 2x^{2}), then (y^\prime=\frac{u^\prime}{u}). Since (u^\prime=-4x), we have (y^\prime=\frac{-4x}{30 - 2x^{2}}=\frac{-2x}{15 - x^{2}}). Set (y^\prime = 0), then (-2x=0), so (x = 0).

Step3: Analyze the first - derivative sign

When (x<0) (in the domain (-\sqrt{15}<x<\sqrt{15})), say (x=-1), (y^\prime=\frac{-2\times(-1)}{15 - 1}=\frac{2}{14}>0). When (x>0) (in the domain (-\sqrt{15}<x<\sqrt{15})), say (x = 1), (y^\prime=\frac{-2\times1}{15 - 1}=\frac{-2}{14}<0). So the function has a local maximum at (x = 0).

Step4: Find the function value at (x = 0)

Substitute (x = 0) into (y=\ln(30 - 2x^{2})), we get (y=\ln(30)). Since the function (y=\ln(30 - 2x^{2})) is continuous on the open interval ((-\sqrt{15},\sqrt{15})) and has only one critical point (x = 0) (where the function changes from increasing to decreasing), this local maximum is also an absolute maximum.

Answer:

A. The absolute maximum value (\ln(30)) occurs at (x = 0).