Question

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

Answer 1

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

We start by finding the first derivative.

`y' = cosx + 2`

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

`0 = cosx + 2`

`-2 = cosx`

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'' = -sinx`

This will equal `0` when `x = pin`. These will be the points of inflection. On `(0, pi)`, the function will be concave down (because the second derivative is negative). On `(pi, 2pi)`, 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!