Intervals in which f(x) is Strictly Increasing/Decreasing Find the intervals in which $f(x)=\sin x + \cos x, 0 \leq x \leq 2 \pi$ is strictly increasing/decreasing.
First I find the derivative $f'(x) = \cos x - \sin x$, then put $f'(x)=0$, getting $\tan x = 1$. The principal solutions of $\tan x = 1$ are $x = \pi/4$ and $x = 5\pi/4$, which gives the intervals $[0,\pi/4)$, $(\pi/4,5\pi/4)$, and $(5\pi/4,2\pi)$. 
After this I am stuck. The book just solves the question by making a table showing the interval, sign of the derivative as positive or negative, and strictly increasing for positive and vice versa.
I just want to know how do they get it positive or negative.It would really help if someone did one for the interval $(\pi/4,5\pi/4)$.
Now, I know the basics of Increasing/Decreasing but our textbook does not mention it thoroughly. I am a student of Class 12(Higher Secondary) so I have a very basic knowledge of Calculus and Trig.
Oh,and if someone wants to know the level of my prescribed school book (and has that much time btw)it is athttp://www.ncert.nic.in/ncerts/l/lemh106.pdf and after page 7, (my question is at page 11)

The function $f'(x)$ is continuous in $[0,2\pi]$. We have $f'(x)=0$ at $x=\frac\pi 4$ and $x=\frac{5\pi}4$.
Then, it can be concluded that $f'(x)$ is either positive for all $x$ in $(\frac\pi4,\frac{5\pi}4)$ or negative for all $x$ in $(\frac\pi4,\frac{5\pi}4)$. This follows from the continuity of $f'(x)$ and the intermediate value theorem.
Now, all that is left to find is whether it takes positive or negative values. This can be done easily by checking the value of $f'(x)$ at any convenient $x$ in the interval. 
For example, at $x=\frac\pi2$, $f'(x)=-1$.
Thus, $f'(x)$ is negative in $(\frac\pi4,\frac{5\pi}4)$ and it can be concluded that $f(x)$ is strictly decreasing in this interval.
