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

identify the coordinates of any local and absolute extreme points and inflection points. graph the function.\n\n$f(x)=\\ln (3 - 5x^{2})$\n\nidentify the coordinates of the local minimum points. select the correct choice below and,\nif necessary, fill in the answer box to complete your choice.\n\na. the local minimum point(s) is/are\n(type an ordered pair. use a comma to separate answers as needed. type an\nexact answer.)\n\nb. there are no local minimum points.\n\nidentify the coordinates of the absolute maximum points. select the correct choice below and,\nif necessary, fill in the answer box to complete your choice.\n\na. the absolute maximum point(s) is/are\n(type an ordered pair. use a comma to separate answers as needed. type an\nexact answer.)\n\nb. there are no absolute maximum points.

identify the coordinates of any local and absolute extreme points and inflection points. graph the function.\n\n$f(x)=\\ln (3 - 5x^{2})$\n\nidentify the coordinates of the local minimum points. select the correct choice below and,\nif necessary, fill in the answer box to complete your choice.\n\na. the local minimum point(s) is/are\n(type an ordered pair. use a comma to separate answers as needed. type an\nexact answer.)\n\nb. there are no local minimum points.\n\nidentify the coordinates of the absolute maximum points. select the correct choice below and,\nif necessary, fill in the answer box to complete your choice.\n\na. the absolute maximum point(s) is/are\n(type an ordered pair. use a comma to separate answers as needed. type an\nexact answer.)\n\nb. there are no absolute maximum points.

Answer

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 (-\sqrt{\frac{3}{5}}<x<\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}). (u^\prime=-10x), so (y^\prime=\frac{-10x}{3 - 5x^{2}}). Set (y^\prime = 0), then (-10x=0), which gives (x = 0).

Step3: Analyze the sign of the first - derivative

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 maximum. (f(0)=\ln(3)).

Step4: Check for local minima

Since the function is increasing on ((-\sqrt{\frac{3}{5}},0)) and decreasing on ((0,\sqrt{\frac{3}{5}})), there are no local minima.

Step5: Check for absolute maxima

Since the function has only one critical point (x = 0) in the domain ((-\sqrt{\frac{3}{5}},\sqrt{\frac{3}{5}})) and the function is continuous on the open interval ((-\sqrt{\frac{3}{5}},\sqrt{\frac{3}{5}})), and (\lim_{x\rightarrow\pm\sqrt{\frac{3}{5}}^{-}}\ln(3 - 5x^{2})=-\infty). The absolute maximum occurs at (x = 0) and (f(0)=\ln(3)).

Answer:

For local minimum points: B. There are no local minimum points. For absolute maximum points: A. The absolute maximum point(s) is/are ((0,\ln(3)))