Question

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

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}

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}When \(x>0\) (in the domain \(-\sqrt{15}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 \(x = 0\).