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 any local minima. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. the local minimum/minima is/are located at
(type an ordered pair. type an exact answer. use a comma to separate answers as needed.)
b. there are no local minima.

identify any local maxima. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. the local maximum/maxima is/are located at
(type an ordered pair. type an exact answer. use a comma to separate answers as needed.)
b. there are no local maxima.

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 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 \(y^\prime=\frac{u^\prime}{u}\).
\(u^\prime=-4x\), so \(y^\prime=\frac{-4x}{30 - 2x^{2}}=\frac{-2x}{15 - x^{2}}\).
Set \(y^\prime = 0\), then \(-2x=0\) gives \(x = 0\).

Step3: Use the first - derivative test

Choose test intervals:

• For \(x\in(-\sqrt{15},0)\), let \(x=-1\), then \(y^\prime=\frac{-2\times(-1)}{15-(-1)^{2}}=\frac{2}{14}>0\).
• For \(x\in(0,\sqrt{15})\), let \(x = 1\), then \(y^\prime=\frac{-2\times1}{15 - 1^{2}}=\frac{-2}{14}<0\).

Since the function changes from increasing to decreasing at \(x = 0\), there is a local maximum at \(x = 0\).
When \(x = 0\), \(y=\ln(30-2\times0^{2})=\ln(30)\).

Step4: Check for local minima

Since the function only changes from increasing to decreasing (no change from decreasing to increasing) in the domain \((-\sqrt{15},\sqrt{15})\), there are no local minima.

Answer:

• Local minima: B. There are no local minima.
• Local maxima: A. The local maximum/maxima is/are located at \((0,\ln(30))\)