# Calculus - Find critical points

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.

## 2 Answers

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.

• Why the downvote? – apnorton Oct 8 '13 at 13:53

$$f'(x) = 1 - 2\cos x = 0 \implies \cos x = \dfrac 12 \implies x = \pi/3 \;\text{or}\; x = 5\pi /3$$ for $x \in [0 , 2\pi]$, and in general $$x = \pm \frac \pi 3 + 2\pi n, \; n \in \mathbb Z$$

ADDED: Note that at $x = \pi/3, \;2\sin x = 2 \sin\left(\frac \pi 3\right) = \sqrt 3 > 0$, and at $x = \frac {5\pi}{3},\;\; 2\sin x = 2 \sin\left(\frac{5\pi}3\right) = -\sqrt 3 < 0$. And in general $$x = \pi/3 + 2\pi n > 0 \implies f''(x) > 0,\quad x = -\pi/3 + 2\pi n \implies f''(x) < 0, \quad n\in \mathbb N$$

Now what does that tell you about when does $f(x) achieve its maximum, and when its minimum? • Technically$-\pi/3$is not equal to$5\pi/3$. – Thomas Oct 8 '13 at 13:48 • I didn't say they were equal. I said they are both solutions. – Namaste Oct 8 '13 at 13:49 • No, they are two different real numbers. Just because the function$f(x) = \cos(x)$takes the same value at two different numbers, don't make the two numbers equal ... – Thomas Oct 8 '13 at 13:50 • If we're talking angles,$-\pi/3$is equivalent to$5\pi/3$. But, in general,$-\pi/3\ne 5\pi/3\$--the first is a negative real number, and the second is a positive real number. That's probably what Thomas is pointing out... – apnorton Oct 8 '13 at 13:52
• @Thomas No offense taken. ;-) I can understand why you'd have thought that, but please be assured, I much prefer to be a colleague, and not a competitor! ;-) – Namaste Oct 8 '13 at 14:09