Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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



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

share|cite|improve this question
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
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 $x \in (5\pi/3, 2\pi]$.

share|cite|improve this answer
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
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
Does that help? Did I answer your question? – amWhy Mar 2 '13 at 22:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.