How do you graph #y=sinx+2x#?
1 Answer
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
#0 = cosx + 2#
#-2 = cosx#
But since
The second derivative can tell us more about concavity and points of inflection.
#y'' = -sinx#
This will equal
In the end, the graph of the function looks very much like this:
Hopefully this helps!