# How do you find local maximum value of f using the first and second derivative tests: f(x)= sinx?

Nov 15, 2016

The max occurs at $x = \frac{\pi}{2}$ where $f \left(x\right) = 1$

#### Explanation:

$1.$ The 1st derivative test: $f ' \left(x\right) = \cos x$

$\cos x = 0$ at $x = \frac{\pi}{2} , \frac{3 \pi}{2} , \text{etc.}$

Make a sign chart for $f ' \left(x\right)$ with a critical point at $\frac{\pi}{2}$. To the left of $\frac{\pi}{2}$, $\cos x$ is positive; to the right, it is negative. Therefore, there is a max at $\frac{\pi}{2}$

$2.$ The 2nd derivative test: ${f}^{' '} \left(x\right) = - \sin \left(x\right)$

At the critical point where $x = \frac{\pi}{2}$, you have

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

Since the second derivative is negative at $x = \frac{\pi}{2}$, the curve is concave down at that point. Therefore, it is a max.