Question

identify the coordinates of any local and absolute extreme points, inflection points, and intervals of concavity. graph the function.

$f(x)=\ln (30 - 2x^{2})$

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 $\ln (30)$ occurs at $x = 0$.
(use a comma to separate answers as needed. type each answer only once.)
b. there is no absolute maximum point.

identify the absolute minimum value and where it occurs. select the correct choice and, if necessary, fill in the answer box to complete your choice.

a. the absolute minimum value occurs at $x =$
(use a comma to separate answers as needed. type each answer only once.)
b. there is no absolute minimum.

Explanation:

Step1: Determine the domain of the function

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

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 \(u^\prime=-4x\), \(y^\prime=\frac{-4x}{30 - 2x^{2}}=\frac{-2x}{15 - x^{2}}\).
Set \(y^\prime = 0\), then \(-2x=0\) gives \(x = 0\).
We also check the behavior as \(x\to\pm\sqrt{15}^{-}\). As \(x\to\pm\sqrt{15}^{-}\), \(y=\ln(30 - 2x^{2})\to-\infty\).

Step3: Analyze the second - derivative

First, rewrite \(y^\prime=\frac{-2x}{15 - x^{2}}\). Using the quotient rule \((\frac{f}{g})^\prime=\frac{f^\prime g - fg^\prime}{g^{2}}\), where \(f=-2x\), \(f^\prime=-2\), \(g = 15 - x^{2}\), \(g^\prime=-2x\).
\(y^{\prime\prime}=\frac{-2(15 - x^{2})-(-2x)(-2x)}{(15 - x^{2})^{2}}=\frac{-30 + 2x^{2}-4x^{2}}{(15 - x^{2})^{2}}=\frac{-30 - 2x^{2}}{(15 - x^{2})^{2}}\).
Since \(y^{\prime\prime}<0\) for all \(x\in(-\sqrt{15},\sqrt{15})\), the function is concave down on \((-\sqrt{15},\sqrt{15})\). There are no inflection points because the concavity does not change.
Since the function \(y = f(x)=\ln(30 - 2x^{2})\) has \(y^\prime = 0\) at \(x = 0\) and \(y^{\prime\prime}(0)=\frac{-30}{15^{2}}<0\), and as \(x\to\pm\sqrt{15}^{-}\), \(y\to-\infty\).

Answer:

For the absolute minimum:
Since as \(x\to\pm\sqrt{15}^{-}\), \(y = \ln(30 - 2x^{2})\to-\infty\), the function has no absolute minimum. So the answer is B. There is no absolute minimum.