# How do you graph y=sinx+2x?

Aug 13, 2017

I'm going to take a calculus approach to this problem.

We start by finding the first derivative.

$y ' = \cos x + 2$

Then there will be critical points where $y ' = 0$.

$0 = \cos x + 2$

$- 2 = \cos x$

But since -1 ≤ cosx ≤ 1, there will be no critical points. The derivative is positive on all $x$, therefore the function is increasing on all of it's domain.

The second derivative can tell us more about concavity and points of inflection.

$y ' ' = - \sin x$

This will equal $0$ when $x = \pi n$. These will be the points of inflection. On $\left(0 , \pi\right)$, the function will be concave down (because the second derivative is negative). On $\left(\pi , 2 \pi\right)$, the function will be concave up (because the second derivative is positive). It will alternate like this to $+$ and $-$ infinity.

In the end, the graph of the function looks very much like this:

Hopefully this helps!