# What are the extrema of f(x)=-sinx-cosx on the interval [0,2pi]?

Nov 8, 2015

Since $f \left(x\right)$ is differentiable everywhere, simply find where $f ' \left(x\right) = 0$

#### Explanation:

$f ' \left(x\right) = \sin \left(x\right) - \cos \left(x\right) = 0$

Solve:

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

Now, either use the unit circle or sketch a graph of both functions to determine where they are equal:

On the interval $\left[0 , 2 \pi\right]$, the two solutions are:

$x = \frac{\pi}{4}$ (minimum) or $\frac{5 \pi}{4}$ (maximum)

hope that helps