13)$\displaystyle f(x)=sin x + cos x 0\leq x \leq 2\pi$

$\displaystyle f'(x)=cosx-sinx = 0$

$\displaystyle x = \frac{\pi}{4} and \frac{5\pi}{4}$

how do I find where the function increase and decrease?

If I put these into a sign chart, I end up with all 0's

also how do I find the concavity?

I get that$\displaystyle f''(x) = -sin x - cos x$

so $\displaystyle x = \frac{7\pi}{4} and \frac{3\pi}{4}$

and for this also, if I plug it into f''(x) for a sign chart, the results will still be 0

is there another method for trig to do these problems?