I need help finding the critical points of this function:
$f(x)=x-2 \sin x $
I found $f'(x)=1-2 \cos x $ and
$f''(x)=2\sin x$
I know the next step is to set $f'(x)=0$ but when I do that I get $x=1.047$.
But looking at the graph I see there is another critical point. How do I obtain this other point?
I also need to find the regions where the graph is concave upward and concave downward.
Best Answer
So you want to solve $$ 0 = f'(x) = 1 - 2\cos(x) $$ which means that $$ \cos(x) = \frac{1}{2}. $$
This happens for example when $x = \frac{\pi}{3}\simeq 1.047$. But remember that $\cos$ is an even function so we also get the solution $x = -\frac{\pi}{3}$. Then remember that $\cos$ is $2\pi$ periodic, so we actually get infinitely many solutions: $$ x = \pm\frac{\pi}{3} + n2\pi, $$ where $n$ is any integer.