1. ## critical points

I apologize. I put this up and then realized the solution. Please delete this.

Find the critical points.
$f(x)=x^3 - 2x - 2 \cos x$
$f'(x)= 3x^2 -2 + 2 \sin x$

So I set the derivatie to 0, to find the critical points, and got:
$3x^2 + 2 sin x - 2 = 0$ ... looks like some kind of spooky quadratic equation.
I know I need to find how those 2 equal 0.
I don't know how, aside from lots of trial and error.

Thank yoU!

edit: solution [SIZE="7"]GRAPH!

2. Originally Posted by Truthbetold
Find the critical points.
$f(x)=x^3 - 2x - 2 \cos x$
$f'(x)= 3x^2 -2 + 2 \sin x$

So I set the derivatie to 0, to find the critical points, and got:
$3x^2 + 2 sin x - 2 = 0$ ... looks like some kind of spooky quadratic equation.
I know I need to find how those 2 equal 0.
I don't know how, aside from lots of trial and error.

Thank yoU!
There's no exact solution that I can see (it's worth testing x = 0, but no it doesn't work). And life's too short for trial and error (besides, there's no exact solution). Any reason why you can't use technology to get a numerical solution to a required level of accuracy ...... Then it's a cinch.