Question

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

$f(x)=\ln (3 - 5x^{2})$

identify the coordinates of the local maximum points. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

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

Explanation:

Step1: Find the domain

For \(y = \ln(3 - 5x^{2})\), the argument of the logarithm must be positive. So \(3-5x^{2}>0\), which gives \(x^{2}<\frac{3}{5}\) or \(x\in(-\sqrt{\frac{3}{5}},\sqrt{\frac{3}{5}})\).

Step2: Differentiate the function

Using the chain - rule, if \(y=\ln(u)\) and \(u = 3-5x^{2}\), then \(y^\prime=\frac{u^\prime}{u}\).
\(u^\prime=-10x\), so \(y^\prime=\frac{-10x}{3 - 5x^{2}}\).

Step3: Find critical points

Set \(y^\prime = 0\), then \(\frac{-10x}{3 - 5x^{2}}=0\). Since the denominator \(3 - 5x^{2}
eq0\) for \(x\in(-\sqrt{\frac{3}{5}},\sqrt{\frac{3}{5}})\), we solve \(-10x = 0\), which gives \(x = 0\).

Step4: Use the second - derivative test

First, find the second derivative. Using the quotient rule \((\frac{f}{g})^\prime=\frac{f^\prime g - fg^\prime}{g^{2}}\), where \(f=-10x\) and \(g = 3-5x^{2}\).
\(f^\prime=-10\) and \(g^\prime=-10x\).
\(y^{\prime\prime}=\frac{-10(3 - 5x^{2})-(-10x)(-10x)}{(3 - 5x^{2})^{2}}=\frac{-30 + 50x^{2}-100x^{2}}{(3 - 5x^{2})^{2}}=\frac{-30 - 50x^{2}}{(3 - 5x^{2})^{2}}\).
When \(x = 0\), \(y^{\prime\prime}=\frac{-30}{9}<0\).

Step5: Find the \(y\) - value at \(x = 0\)

When \(x = 0\), \(y=\ln(3-5\times0^{2})=\ln(3)\).

Answer:

A. The local maximum point(s) is/are \((0,\ln(3))\)