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

Question

Isaiah Allen · Accepted Answer

Since #f(x)# is differentiable everywhere, simply find where #f'(x)=0#

#f'(x)=sin(x)-cos(x)=0#

Solve:

#sin(x)=cos(x)#

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

On the interval #[0,2pi]#, the two solutions are:

#x=pi/4# (minimum) or #(5pi)/4# (maximum)

hope that helps

Charles Anctil · Answer

undefined To find the extrema of $ f(x) = -\sin(x) - \cos(x) $ on the interval $[0, 2\pi]$, we first find the critical points by setting the derivative of $ f(x) $ equal to zero:

$ f'(x) = -\cos(x) + \sin(x) $

Setting $ f'(x) = 0 $:

$ -\cos(x) + \sin(x) = 0 $

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

Solving for $ x $:

$ x = \frac{\pi}{4}, \frac{5\pi}{4} $

Now, we evaluate the function at the critical points and at the endpoints of the interval $[0, 2\pi]$:

$ f(0) = -\sin(0) - \cos(0) = -1 $

$ f\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = -\sqrt{2} $

$ f\left(\frac{5\pi}{4}\right) = -\sin\left(\frac{5\pi}{4}\right) - \cos\left(\frac{5\pi}{4}\right) = -\sqrt{2} $

$ f(2\pi) = -\sin(2\pi) - \cos(2\pi) = -1 $

The extrema occur at the critical points $ \frac{\pi}{4} $ and $ \frac{5\pi}{4} $, where the function takes on the value $ -\sqrt{2} $. The minimum value of the function is $ -\sqrt{2} $, and the maximum value is $ -1 $.