## anonymous 3 years ago At what value of x does the local max of f(x) occur?

1. anonymous

$f(x)=\int\limits_{0}^{x}\frac{ t^2-1 }{ 1+\cos^2(t) }dt$

2. anonymous

So they gave you the derivative... When it is 0?

3. anonymous

Also, when is it undefined?

4. anonymous

It seems it is always defined.

5. anonymous

so it's zero when x=0, right?

6. zepdrix

Ummmmmm I think what we want to do is use the Fundamental Theorem of Calculus, Part 1 for this problem. Which states,$\large \frac{d}{dx} \int\limits_0^{x}f(t)dt \qquad = \qquad f(x)$ For this problem, we're given $$\large f(x)$$. We can find a maximum by first finding critical points the function. Remember how to do that? Critical points occur when $$\large f'(x)=0$$. $\large f(x)=\int\limits\limits_{0}^{x}\frac{ t^2-1 }{ 1+\cos^2(t) }dt$ Applying the FTC, Part 1 gives us,$\large f'(x)=\frac{x^2-1}{1+\cos^2(x)}$

7. anonymous

Ok yeah I'm following that now

8. zepdrix

Setting it equal to zero, $\large 0=\frac{x^2-1}{1+\cos^2(x)}$ We should be able to find some critical points, and then by some method we can determine which point(s) are max/min or whatever.

9. anonymous

so the critical points are -1 and 1

10. zepdrix

cool :) Remember how to determine if each point is a max or min?

11. anonymous

using the 2nd derivative test so the max is at -1

12. zepdrix

Yah that's what I'm coming up with also. Cool!

13. anonymous

THANKS!

14. zepdrix

heh np c: