Given f(x) = ∫(low: 0, up:x) (t^2 - 25 / 1 + cos^2(t))dt?

Question

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

x = ________.

Answer 1

By the Fundamental Theorem of Calculus,
f'(x) = x^2 - 25 / 1 + cos^2 x
- since you haven't supplied parentheses I have to guess at the precedence here; I'll assume this is
(x^2 - 25) / (1 + cos^2 x)
but it could also realistically be x^2 - (25 / (1 + cos^2 x)) - but I don't think so, because you can't solve this analytically.
So we want (x^2 - 25) / (1 + cos^2 x) = 0 - note that the denominator is always >= 1. So f'(x) = 0 at x = ±5, and we can see that at x = -5 f'(x) changes from positive to negative (MAX) and at x = 5 f'(x) changes from negative to positive (MIN). So the local max occurs at x = 5.

Answer 2

The local max of f(x) occurs at -5