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.
1 Answer
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 $x \in [0, 2\pi]$.
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 $x \in (5\pi/3, 2\pi]$.
