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

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

2. wio Group Title

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

3. wio Group Title

Also, when is it undefined?

4. wio Group Title

It seems it is always defined.

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

6. zepdrix Group Title

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 Group Title

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 Group Title

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 Group Title

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