Find min and maxima

Question

Find local min and maxima of $ \sin(x^3)$ on the interval $]-2,2[$.

I take the derivative and get:

$$3x^2 \cdot \cos (x^3)$$

I set this equal to zero and get $$x^3 = \cos^{-1}(0)$$

$$ \Rightarrow x^3 = \frac{\pi}{2}+k \cdot \pi$$

I'm not sure what the easiest way to proceed from here would be? Taking the third root confuses me more than just keeping it on this format..I think.

Answer 1

Setting the derivative equal to zero gives you $$3x^2\cdot\cos(x^3)=0$$ and this is zero only when $x=0$ or $\cos(x^3)=0$.

$x=0$ can't be a local minimum, because $\sin(x^3)$ is an odd function, so it remains to solve $\cos(x^3)=0$ on $]\!-\!2,2[$.

Note that $\cos^{-1} t$ has a specific meaning, it's the inverse of $\cos t$ restricted to $[0,\pi]$. So just say that $\cos t=0$ for $\displaystyle t=\frac\pi2+k\pi$, where $k\in\mathbb Z$ and you want to solve $$x^3=\frac\pi2+k\pi$$ on $]\!-\!2,2[$. But $x^3$ is one-to-one and transforms that interval into $]\!-\!8,8[$. Can you find all values of $k$ such that $\displaystyle\frac\pi2+k\pi\in\ ]\!-\!8,8[$ ?

Answer 2

Because it is $ \sin$ of something ($ \sin(X)$) the minimum is always -1 and the maximum is +1.

As you can see in this interval it has exactly 3 minimums and 3 maximums if the interval is closed and 2 if the interval is open like in your case.

Answer 3

There is no need to compute any derivative: the expression is maximum when $$ x^3 = \frac\pi 2 + n\pi, n\in\Bbb Z. $$