 # Figuring out when f(x)=sin(x^2) is increasing and decreasing. Silvina2b 2022-08-11 Answered
Figuring out when $f\left(x\right)=\mathrm{sin}\left({x}^{2}\right)$ is increasing and decreasing
Regarding the function $f\left(x\right)=\mathrm{sin}\left({x}^{2}\right)$, I'm supposed to figure out when it is increasing/decreasing.
So far, I've found the derivative to be ${f}^{\prime }\left(x\right)=2x\mathrm{cos}\left({x}^{2}\right)$.
So long as I can solve the inequality $\mathrm{cos}\left({x}^{2}\right)>0$. I can figure the rest out, but this is where I'm stuck.
I've narrowed it down to $-\frac{\pi }{2}<{x}^{2}<\frac{\pi }{2}$ meaning .
Then I get $x<±\sqrt{\frac{\pi }{2}±n2\pi }$.
Am I on the right track here? I can't seem to find a path from here.
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Addison Herman
Step 1
You're on the right track.
The "points that matter" in partitioning the increasing/decreasing intervals are the points for which ${f}^{\prime }\left(x\right)=0$. This happens when either $2x=0$ or $\mathrm{cos}\left({x}^{2}\right)=0$.
The function is increasing when ${f}^{\prime }\left(x\right)>0$, and decreasing when ${f}^{\prime }\left(x\right)<0$.
Consider each part of the derivative separately.
The linear part of the derivative, 2x, is positive when $x>0$, and negative when $x<0$.
Step 2
The cosine function part of the derivative, $\mathrm{cos}\left({x}^{2}\right)$, is positive when its argument is on one the open intervals
$\left(2\pi n-\pi /2,2\pi n+\pi /2\right),n\in \mathbb{Z},$
and negative when it's on one the open intervals $\left(2\pi n+\pi /2,2\pi n+3\pi /2\right),n\in \mathbb{Z}.$
The next steps would involve looking at when the product of these two parts changes sign. Those points would define your increasing/decreasing intervals.

We have step-by-step solutions for your answer!