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$f(x) = \sin(x)+ \cos(x) $ for $ 0≤x≤2\pi$

I have to do the following

1) Find the intervals on which $f$ is increasing or decreasing

2) Find the local maximum and minimum values of $f$

3) Find the intervals of concavity and the inflections points

I got until this point when trying to solve problem 1: $\tan(x) = 1$. The next step in the manual says that $x = \pi/4$ or $5\pi/4$. How did they get that $x = \pi/4$ or $5\pi/4$?

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My question is how π/4 or 5π/4 = x? – user44816 Nov 20 '12 at 17:57
The sine, cosine, and probably tangent of some "special" angles should be memorized. You may remember that the tangent of the $45^\circ$ angle is $1$. There is also an angle in the third quadrant whose tan is $1$. It is obtained by adding $180^\circ$ to the $45^\circ$ angle. The fact that the tangent of the $45^\circ$ angle is $1$ comes from looking at the isosceles right-angled triangle. – André Nicolas Nov 20 '12 at 17:58
I would prefer not to use calculus for this question - if we write $\sin(x)+\cos(x)$ as $\sqrt{2}\sin(x+\frac{\pi}{4})$, then all the answers are obvious from basic facts of the graph. – Daniel Littlewood Nov 21 '12 at 20:59
up vote 3 down vote accepted

We solve for $0\leq x\leq 2\pi$. $\tan x=1$ if and only if $x=\tan^{-1} 1=\frac{\pi}{4}$ or $x=\pi+\tan^{-1} 1=\frac{5\pi}{4}$. ($\frac{\pi}{4}$ is the principal value and the tangent is positive in the first and third quadrants.)

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(1) $f$ is increasing or decreasing accordingly as $f'(x)>0$ or $<0$

Now, $f'(x)=\cos x-\sin x=\sqrt2\cos(x+\frac \pi 4)$

$f'(x)>0$ if $\cos(x+\frac \pi 4)>0$ if $ x+\frac \pi 4$ lies in the 1st or 4th quadrant.

As,$0\le x\le 2\pi,$ for $f'(x)>0, 0\le x<\frac \pi 4$ or $\frac{3\pi}2-\frac \pi4<x\le 2\pi$

Similarly for $f'(x)<0$

(2)For the maxima/minima, $f'(x)=0\implies \tan x=1=\tan \frac \pi 4 \implies x=m\pi+\frac \pi 4 $ where $m$ is any integer.

$f''(x)=-(\sin x+\cos x)=-\sqrt2\cos(x-\frac \pi 4)$

So, $f''(m\pi+\frac \pi 4)= -\sqrt2\cos m\pi$ which is $<0$ if $m$ is even$=2r$(say) where $r$ is any integer

So,the local maximum of $f(x)$ is at $x=2r\pi+\frac \pi 4$

As $0\le x\le 2\pi,$ for local maximum of $f(x),x=\frac \pi 4$

$f_{max}=f(\frac \pi 4)=\sqrt 2$

Similarly, for local minimum.

(3) Using this, $f(x)$ is concave up if $f''(x) > 0$

So, we need $\cos(x-\frac \pi 4)<0$ i.e., $x-\frac \pi 4$ will lie in the 2nd or in the 3rd quadrant.

$\frac \pi2<x-\frac \pi 4< \frac {3\pi}2\implies \frac \pi2+\frac \pi 4<x< \frac {3\pi}2+\frac \pi 4$ as $0\le x\le 2\pi$

Similar for the concave down if $f''(x) < 0$

For the point of inflexion, $f''(x)=0$ or $f''(x)$ does not exist.

Here clearly $f^n(x)$ exists for $n\ge 0$

So we need $f''(x)=0\implies \sin x+\cos x=0\implies \tan x=-1=\tan (-\frac{\pi}4)\implies x=s\pi-\frac{\pi}4$ where $s$ is any integer

As $0\le x\le 2\pi,x=\pi-\frac{\pi}4,2\pi-\frac{\pi}4$ for the point of inflexion.

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