# How do you use the second derivative test to find the local extrema for f(x)=sin(x)?

Apr 2, 2018

Local maxima: $\left(\frac{\pi}{2} + 2 n \pi , 1\right)$ Local minima: $\left(\frac{3 \pi}{2} + 2 n \pi , - 1\right)$

#### Explanation:

Differentiate the function, equate to zero, and solve.

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

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

$\cos x = 0 \to x = \frac{\pi}{2} + n \pi$ where $n$ is an integer.

Now, the second derivative tells us that if $a$ is a critical value of $f \left(x\right)$ (a value which causes the derivative to go to zero), if $f ' ' \left(a\right) > 0 ,$ there is a minimum at $\left(a , f \left(a\right)\right) ,$ and if $f ' ' \left(a\right) < 0 ,$ then there is a maximum at $\left(a , f \left(a\right)\right)$.

As seen above, this function has infinitely many critical values due to the periodic nature of the function. But let's take $x = \frac{\pi}{2} , \frac{3 \pi}{2}$ and test them in $f ' ' \left(x\right) .$

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

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

So, at $x = \frac{\pi}{2} + 2 n \pi$ there are maxima.

$f ' ' \left(\frac{3 \pi}{2}\right) = - \sin \left(\frac{3 \pi}{2}\right) = - \left(- 1\right) = 1 > 0$

So, at $x = \frac{3 \pi}{2} + 2 n \pi ,$ there are minima.

$f \left(\frac{\pi}{2}\right) = 1$ so the local maxima are $\left(\frac{\pi}{2} + 2 n \pi , 1\right)$

$f \left(\frac{3 \pi}{2}\right) = - 1$ so there are local minima at $\left(\frac{3 \pi}{2} + 2 n \pi , - 1\right)$