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## heradog Group Title At what value of x does the local max of f(x) occur? one year ago one year ago

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1. heradog

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

2. wio

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

3. wio

Also, when is it undefined?

4. wio

It seems it is always defined.

5. heradog

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. heradog

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. heradog

so the critical points are -1 and 1

10. zepdrix

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

11. heradog

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. heradog

THANKS!

14. zepdrix

heh np c: