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

1 Answer
Mar 3, 2018

See below

Explanation:

#y = sinx+x#

Note that #y# is defined #forall x in RR#

Now, since #-1<= sinx <= +1#

Then, #lim_(x->-oo) y = -oo and lim_(x->+oo) y = +oo#

To find the critical points of #y# we will set the first differential #y'=0# and test with the second differential #y''#.
(You will need to refer to Calculus here)

#y' = cosx+1 = 0#

#cosx =-1 -> x = npi: forall n in ZZ#

#y'' = -sinx#

Now, #-sinx = 0 # for #x = npi:forall n in ZZ#

So, #y# has inflection points for #x = npi: forall n in ZZ#

And, as #absx# increases #y->x#

We can see these results on the graph of #y# below.
(If you zoom out the graph #-> y=x#)

graph{sinx+x [-65.83, 65.87, -32.9, 32.9]}