# Find the absolute extrema of the given function f(x)= sinx+cosx on interval [0,2pi]?

Oct 16, 2015

${f}_{\max} = \sqrt{2}$
${f}_{\min} = - \sqrt{2}$

#### Explanation:

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

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

$\cos \frac{x}{\cos} x - \sin \frac{x}{\cos} x = 0 \wedge \cos x \ne 0$

$\tan x = 1 \wedge \cos x \ne 0$

$x = \frac{\pi}{4} + k \pi \wedge x \ne \frac{\pi}{2} + m \pi$

$x = \frac{\pi}{4} + k \pi$

On $\left[0 , 2 \pi\right]$ :

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

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

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

${f}_{\max} = \sqrt{2}$
${f}_{\min} = - \sqrt{2}$