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 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 $(0,\ln(30))$.

(type an ordered pair. type an exact answer. use a comma to separate answers as needed.)

b. there are no local maxima.

identify any inflection points. select the correct choice and, if necessary, fill in the answer box to complete your choice.

a. the inflection point(s) is/are at

(type an ordered pair. type an exact answer. use a comma to separate answers as needed.)

b. there are no inflection points

Explanation:

Step1: Find the first - derivative

Use the chain rule. If \(y = \ln(u)\) and \(u=30 - 2x^{2}\), then \(y^\prime=\frac{u^\prime}{u}\).
\(u^\prime=-4x\), so \(f^\prime(x)=\frac{-4x}{30 - 2x^{2}}=\frac{-2x}{15 - x^{2}}\), and the domain of \(f(x)\) is \(30-2x^{2}>0\), i.e., \(-\sqrt{15}Set \(f^\prime(x) = 0\), then \(-2x=0\), \(x = 0\).

Step2: Find the second - derivative

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Here \(u=-2x\), \(u^\prime=-2\), \(v = 15 - x^{2}\), \(v^\prime=-2x\).
\(f^{\prime\prime}(x)=\frac{-2(15 - x^{2})-(-2x)(-2x)}{(15 - x^{2})^{2}}=\frac{-30 + 2x^{2}-4x^{2}}{(15 - x^{2})^{2}}=\frac{-30 - 2x^{2}}{(15 - x^{2})^{2}}\)
Set \(f^{\prime\prime}(x)=0\), \(-30 - 2x^{2}=0\), \(x^{2}=-15\) (no real solutions).

Answer:

B. There are no inflection points.