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How would I find the f interval of the following trigonometric function.

$f(x)=x-2 \sin(x)$ defined on $[0,2\pi]$

I did $f'(x)=1-2 \cos(x)$

$\cos(x)=\frac{1}{2}$

$x=\frac{\pi}{3},\frac{5\pi}{3}$

But when I graph it on wolfram alpha there seem to be more zeroe where it increases and decreases.

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By an analysis of your derivative, the function decreases in the interval $[0,\pi/3]$, then increases in $[\pi/3,5\pi/3]$, then decreases in $[5\pi/3,2\pi]$. –  André Nicolas Mar 2 '13 at 21:50
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1 Answer

up vote 1 down vote accepted

No worries, you need only determine when $f(x)$ is increasing or decreasing on for $x \in [0, 2\pi]$

Test between $x = 0$ and $x \lt \pi/3$ (the derivative is negative there).

Test between $x > \pi/3$ and $x < 5\pi/3$, (the derivative is positive there) and then

test the values between $x > 5\pi/3$ and $x\leq 2\pi$ (where the $f'(x)\lt 0$).

The other "zeros" you see occur outside of the interval $x \in [0, 2\pi]$.

enter image description here

So, you need only worry $f(x)$ on the interval $[0, 2\pi]$.

$f(x)$ is decreasing on the intervals $x\in [0, \pi/3)$ and increases on $x\in (\pi/3, 5\pi/3)$, then decreases on the interval $x \in (5\pi/3, 2\pi]$.

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yes that makes sense because it has to be between zero and 2 pi. –  Fernando Martinez Mar 2 '13 at 21:53
    
Yes, on your earlier trig function, as well, we can restrict the intervals of where the function was increasing to $x \in [0, 2\pi]$. Good question. –  amWhy Mar 2 '13 at 21:57
    
I have a quick question when you want to plug in a point to test say 50 degrees do you plug it into the original function not the derivative? –  Fernando Martinez Mar 2 '13 at 22:04
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To test a point $x$ for whether it is increasing or decreasing, you plug it into the derivative: if $f'(x) > 0$ it is increasing there, if $f(x) \lt 0$, it is decreasing there. To find the actual function value at point x, then plug it into the original function. –  amWhy Mar 2 '13 at 22:06
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Does that help? Did I answer your question? –  amWhy Mar 2 '13 at 22:09
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