Question

The minimum value of f(x) = sin x in $\left[\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ is ____________________.

Answer 1

The given function is $f\left(x\right)=\sin x$, $x\in \left[\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$.

$f\left(x\right)=\sin x$

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=\cos x$

For maxima or minima,

$f\text{'}\left(x\right)=0$

$\Rightarrow \cos x=0$

$\Rightarrow x=-\frac{\mathrm{\pi }}{2}$ or $x=\frac{\mathrm{\pi }}{2}$, for all $x\in \left[\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$

Now,

$f\text{'}\text{'}\left(x\right)=-\sin x$

At $x=\frac{\mathrm{\pi }}{2}$, we have

$f\text{'}\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=-\sin \frac{\mathrm{\pi }}{2}=-1<0$

So, $x=\frac{\mathrm{\pi }}{2}$ is the point of local maximum of f(x).

At $x=-\frac{\mathrm{\pi }}{2}$, we have

$f\text{'}\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=-\sin \left(-\frac{\mathrm{\pi }}{2}\right)=\sin \frac{\mathrm{\pi }}{2}=1>0$ [sin(−θ) = −sinθ]

So, $x=-\frac{\mathrm{\pi }}{2}$ is the point of local minimum of f(x).

∴ Minimum value of f(x) = $f\left(-\frac{\mathrm{\pi }}{2}\right)=\sin \left(-\frac{\mathrm{\pi }}{2}\right)=-\sin \frac{\mathrm{\pi }}{2}=-1$

Thus, the minimum value of f(x) = sinx in $\left[\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ is −1.


The minimum value of f(x) = sin x in $\left[\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ is ___−1___.