Graphs of derivative sin functions I am trying to find the intervals on which f is increasing or decreasing, local min and max, and concavity and inflextion points for $f(x)=\sin x+\cos x$ on the interval $[0,\pi]$.
I know at $\pi/4$ the derivative will equal zero. So that gives me my critical numbers, positive and negative $\pi/4$ so now I need to find the intervals which is not making any sense to me, I thought they could only change at critical numbers but $\pi$ and $2\pi$ are different values. I am getting a positive for $2\pi$ and a negtive for $\pi$. How can this happen if the only critical number is $\pi/4$?
 A: $f(x)=\sin(x)+\cos(x)$
$f'(x)=\cos(x)-\sin(x)$
critical points are when $f'(x)=0$: 
i.e, at:
$\cos(x)=\sin(x)$ which can be satisfied by the values of x such as:
...,$-7{\pi}/4$ , $-3{\pi}/4$, ${\pi}/4$, $3{\pi}/4$,...
now, you need to examine the second derivative's sign at the above points:
$f''(x)=-\sin(x)-\cos(x)$
at $-7{\pi}/4$ , $f''(x)$ is (-) --> Local Max.
at $-3{\pi}/4$,   $f''(x)$ is (+) --> Local Min.
at ${\pi}/4$ , $f''(x)$ is (-) --> Local Max.
at $3{\pi}/4$,   $f''(x)$ is (+) --> Local Min.
The link (example) may help.
also, These plots may help:

A: In short, $\pi / 4$ is not the only critical value. In general, we know $\sin$ and $\cos$ are periodic, so we expect infinitely many critical values. We also know that if $\sin a = 0$, then $\sin (a + \pi) = 0$.
So when you take the derivative, you get $f'(x) = \cos x - \sin x$. You look for critical points - so you check when $\sin x = \cos x$, and you get when they're both positive (which you got), and when they're both negative (which you did not).
Note that $- \pi / 4$ is not a solution.
