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

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

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.

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,$\large f'(x)=\frac{x^2-1}{1+\cos^2(x)}$

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.

so the critical points are -1 and 1

10. zepdrix

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

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

12. zepdrix

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