Question

identify the coordinates of any local and absolute extreme points and inflection points. graph the function.
f(x)=\ln(3 - 5x^{2})

Explanation:

Step1: Determine the domain of the function

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

Step2: Find the first derivative

Using the chain rule, if \(y=\ln(u)\) and \(u = 3-5x^{2}\), then \(y^\prime=\frac{u^\prime}{u}\).
Since \(u^\prime=-10x\), the first derivative \(y^\prime=\frac{-10x}{3 - 5x^{2}}\).
Set \(y^\prime = 0\), then \(-10x=0\) gives \(x = 0\).
When \(x\in(-\sqrt{\frac{3}{5}},0)\), \(y^\prime>0\) (function is increasing). When \(x\in(0,\sqrt{\frac{3}{5}})\), \(y^\prime<0\) (function is decreasing). So, \(x = 0\) is a local (and in this case, absolute) maximum.

Step3: Analyze the behavior as \(x\to\pm\sqrt{\frac{3}{5}}\)

As \(x\to\pm\sqrt{\frac{3}{5}}\), \(3 - 5x^{2}\to0^{+}\), and \(\lim_{x\to\pm\sqrt{\frac{3}{5}}}\ln(3 - 5x^{2})=-\infty\).

Answer:

Since the function \(y = \ln(3 - 5x^{2})\) approaches \(-\infty\) as \(x\) approaches the endpoints of its domain \(x=\pm\sqrt{\frac{3}{5}}\), there are no absolute minimum points. So, the answer is B. There are no absolute minimum points.