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

1 Answer
Noah G
Aug 13, 2017

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:
enter image source here

Hopefully this helps!

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