Question

At what value of x does the local max of f(x) occur?

Answer 1

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

Answer 2

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

Answer 3

Also, when is it undefined?

Answer 4

It seems it is always defined.

Answer 5

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

Answer 6

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

Answer 7

Ok yeah I'm following that now

Answer 8

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.

Answer 9

so the critical points are -1 and 1

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

THANKS!

Answer 14

heh np c: