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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)

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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.

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  • $\begingroup$ That was helpful but If I take x=π/2 at tan(x) won't it be undefined.?Or should I take it at cos(x)-sin(x).That clarification would be nice. $\endgroup$ – Ishan Taneja Feb 11 '16 at 18:27
  • $\begingroup$ @IshanTaneja There is no $\tan x$ in the expression for $f'(x)$, and the function is well defined and continuous at $\frac\pi 2$, so it is fine. You can take any value in the interval, say, $\frac\pi 3$, that you want. $\endgroup$ – GoodDeeds Feb 11 '16 at 18:30
  • $\begingroup$ Oh right,that was for the critical point,anyway thanks.And also would I need to do the calculation of any convenient x for each interval ? $\endgroup$ – Ishan Taneja Feb 11 '16 at 18:32
  • $\begingroup$ @IshanTaneja That was a convenient method in this case, and you can do it easily for each interval. In general, you could consider using the second derivative test to determine the points of local maximum and minimum. The continuous function would be strictly increasing from a point of local minimum to the next adjacent point of local maximum. $\endgroup$ – GoodDeeds Feb 11 '16 at 18:34
  • $\begingroup$ Using the second derivative for Increasing and Decreasing (both together I mean)is not in my syllabus so I will have to make do with that.:) $\endgroup$ – Ishan Taneja Feb 11 '16 at 18:36

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