Regarding the function $f(x) = \sin(x^2)$, I'm supposed to figure out when it is increasing/decreasing.
So far, I've found the derivative to be $f'(x) = 2x\cos(x^2)$.
So long as I can solve the inequality $\cos(x^2) > 0$ I can figure the rest out, but this is where I'm stuck.
I've narrowed it down to $-\frac\pi2 < x^2 < \frac\pi2$ meaning $x^2 < \frac\pi2 \pm n2\pi, \ \ n\in \mathbb N$
Then I get $x < \pm \sqrt{\frac\pi2 \pm n2\pi}$
Am I on the right track here? I can't seem to find a path from here.
Best Answer
You're on the right track.
The "points that matter" in partitioning the increasing/decreasing intervals are the points for which $f'(x) = 0$. This happens when either $2x = 0$ or $\cos(x^2) = 0$.
The function is increasing when $f'(x) > 0$, and decreasing when $f'(x) < 0$.
Consider each part of the derivative separately.
The linear part of the derivative, $2x$, is positive when $x > 0$, and negative when $x < 0$.
The cosine function part of the derivative, $\cos(x^2)$, is positive when its argument is on one the open intervals
$$(2 \pi n - \pi / 2, 2 \pi n + \pi / 2), n \in \mathbb{Z},$$
and negative when it's on one the open intervals
$$(2 \pi n + \pi / 2, 2 \pi n + 3\pi / 2), n \in \mathbb{Z}.$$
The next steps would involve looking at when the product of these two parts changes sign. Those points would define your increasing/decreasing intervals.