Find out the local maxima/minima of $\sin x$ where $x\in\left[\dfrac{\pi}{2},\dfrac{3\pi}{2}\right]$

Question

Find out the local maxima/minima of $\sin x$ where $x\in\left[\dfrac{\pi}{2},\dfrac{3\pi}{2}\right]$

My attempt is as follows:-

$$f'(x)=\cos x$$

So getting local minima/maxima at endpoints, is it fine to declare endpoints as minima/maxima?

Answer 1

You have to also look at the sign of the second derivative in those points to establish if they are local minima or maxima points.

However, there is no need for derivatives if you just know the graph of $\sin x$. It's pretty clear that $\sin $ is decreasing over $\left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right]$, so:

$$\sin \frac{\pi}{2} \geq \sin x \geq \sin \frac{3\pi}{2}\Rightarrow 1\geq \sin x\geq -1$$

Answer 2

Instead of using a derivative, note that $-1 \le \sin(x) \le 1$, with it being $-1$ at $x = \frac{3\pi}{2}$ and $1$ at $x = \frac{\pi}{2}$. Thus, the local minima/maxima of $\sin(x)$ with $x \in\left[\dfrac{\pi}{2},\dfrac{3\pi}{2}\right]$ are $-1$ and $1$, respectively.

As for using extrema at the end points, as shortmanikos's comment and Doug M's comment state, it's fine since you have a closed interval of $\left[\dfrac{\pi}{2},\dfrac{3\pi}{2}\right]$, but be careful if using the derivative as it may not necessarily be $0$ at those points.