# Finding extrema based on a function

• Nov 23rd 2009, 05:54 PM
Chris22
Let f be the function given by $\displaystyle f(x) = 3ln(x^2+2)-2x$ with domain [-2,4].

a) Find the coordinate of each relative maximum and minimum point of f. Explain your answer.

I understand that I have to find the first and second derivative.
For the first derivative I got $\displaystyle \frac {-2x^3+2x}{x^2+2}$

For the second derivative I got $\displaystyle \frac{-2(x^4+7x^2+2)}{(x^2+2)^2}$
I was just wondering if that was right.

I also had to find the x-coordinate of each point of inflection. Am I right when I say that i need to do sign analysis for the second derivative to find the points of inflection?

Thanks,
-Chris
• Nov 23rd 2009, 07:17 PM
skeeter
Quote:

Originally Posted by Chris22
Let f be the function given by $\displaystyle f(x) = 3ln(x^2+2)-2x$ with domain [-2,4].

a) Find the coordinate of each relative maximum and minimum point of f. Explain your answer.

I understand that I have to find the first and second derivative.
For the first derivative I got $\displaystyle \frac {-2x^3+2x}{x^2+2}$

For the second derivative I got $\displaystyle \frac{-2(x^4+7x^2+2)}{(x^2+2)^2}$
I was just wondering if that was right.

I also had to find the x-coordinate of each point of inflection. Am I right when I say that i need to do sign analysis for the second derivative to find the points of inflection?

Thanks,
-Chris

$\displaystyle f'(x) = \frac{6x}{x^2+2} - 2 = \frac{-2(x^2-3x+2)}{x^2+2}$