Question

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

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

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

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 \(-\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 \(x = 0\).
We can also use the sign - chart for \(y^\prime\).
Let's test intervals:

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

Since the function changes from increasing (\(y^\prime>0\)) to decreasing (\(y^\prime<0\)) at \(x = 0\), \(x = 0\) is a local maximum.

Step3: Analyze for local minima

Since the function has only one critical point \(x = 0\) (from the first - derivative test) and it is a local maximum, there are no local minima.

Answer:

B. There are no local minima.