Question

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

Answer 1

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]}