Question

$f\left(x\right)=\sin x\text{has a local minima at}x=\frac{3\pi }{2}$.

• A. True
• B. False

Answer 1

The correct option is A

True

From the definition of local minimum, we know that it is a point which has lower output compared to its neighborhood. In other words f(x) will have a local minimum at $x=a$ if $f\left(a\right)<f(a-h)$ and $f\left(a\right)<f(a+h)$ where h is an infinitesimal positive number.

Let’s find $f\left(a\right),f(a-h)$ and $f(a+h)$.

$f\left(\frac{3\pi }{2}\right)=\text{sin}\left(\frac{3\pi }{2}\right)=-1f(\frac{3\pi }{2}-h)=\text{sin}(\frac{3\pi }{2}-h)=\text{sin}(\pi +\frac{\pi }{2}-h)=-\text{sin}(\frac{\pi }{2}-h)$

We know $\sin \left(\frac{\pi }{2}\right)=1$ and $\text{sin}(\frac{\pi }{2}-h)$ will be slightly less than 1.

We can write it as $-1+h$

$-\sin (\frac{\pi }{2}-h)=-(1-h)=-1+h$

So, $f(\frac{3\pi }{2}-h)$ will be slightly greater than - 1.

Now, let’s find $f(\frac{3\pi }{2}+h)$

$\sin (\frac{3\pi }{2}+h)=\sin (\pi +\frac{\pi }{2}+h)=-\sin (\frac{\pi }{2}+h)$

Here, sine will again be slightly less than 1 as sine has local maximum at $x=\frac{\pi }{2}$.

We can write it as $-1+h$ again.

And in the same way $\sin (\frac{3\pi }{2}+h)=-1+h$

Which is slightly greater than -1.

$\mathrm{Here}\mathrm{we}\mathrm{find}\mathrm{that}f\left(\frac{3\pi }{2}\right)<f\left(\frac{3\pi }{2}-h\right)\mathrm{and}f\left(\frac{3\pi }{2}\right)<f\left(\frac{3\pi }{2}+h\right)$

Hence we can say that sin(x) has a minimum at $x=\frac{3\pi }{2}$