f(x)=sinx has a local minima at x=3π2.
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(a)<f(a−h) and f(a)<f(a+h) where h is an infinitesimal positive number.
Let’s find f(a),f(a−h) and f(a+h).
f(3π2)=sin(3π2)=−1f(3π2−h)=sin(3π2−h)=sin(π+π2−h)=−sin(π2−h)
We know sin(π2)=1 and sin(π2−h) will be slightly less than 1.
We can write it as −1+h
−sin(π2−h)=−(1−h)=−1+h
So, f(3π2−h) will be slightly greater than - 1.
Now, let’s find f(3π2+h)
sin(3π2+h)=sin(π+π2+h)=−sin(π2+h)
Here, sine will again be slightly less than 1 as sine has local maximum at x=π2.
We can write it as −1+h again.
And in the same way sin(3π2+h)=−1+h
Which is slightly greater than -1.
Here we find thatf(3π2)<f(3π2−h) and f(3π2)<f(3π2+h)
Hence we can say that sin(x) has a minimum at x=3π2