# How do you find the maximum of f(x) = 2sin(x^2)?

Mar 6, 2015

The maximum value of $f \left(x\right) = 2 \sin \left({x}^{2}\right)$ is $2$.

This is an example of a kind of problem sometimes asked in a calculus class. It is a kind of a trap question. You do not need calculus to answer this question.

The maximum value of the sine function is $1$. This happens when ${x}^{2} = \frac{\pi}{2} + 2 \pi k$ for any integer $k$.

I also know that ${x}^{2} = \frac{\pi}{2}$ when $x = \sqrt{\frac{\pi}{2}}$.

So the maximum value of this function is $2$.

If you need to know where else the maximum occurs, then you need to solve: ${x}^{2} = \frac{\pi}{2} + 2 \pi k = \frac{\pi}{2} \left(1 + 4 k\right)$ for any integer $k$.

The solutions are $x = \pm \sqrt{\frac{\pi}{2}} \sqrt{1 + 4 k}$ which might be easier to read written: $x = \pm \sqrt{1 + 4 k} \sqrt{\frac{\pi}{2}}$ for any integer k.