Question

Question #92256

Answer 1

See explanation

Explanation:

Break this into two parts, firstly the inner part:

`e^x`

This is positive and increasing for all real numbers and goes from 0 to `oo` as `x` goes from `-oo` to `oo`

The we have:

`arctan(u)`

The has a right horizontal asymptote at `y=pi/2`. Going from `u=0 rarr oo`, at `u=0` this function is positive and increasing over this domain, takes a value of 0 at `u=0`, a value of `pi/4` at `u=1` and a value of `pi/2` at `u=oo`.

These points therefore get pulled to `x=-oo,0,oo` respectively and we end up with a graph looking like this as a result:

graph{arctan(e^x) [-10, 10, -1.5, 3]}

Which is the positive part of the `arctan` function stretch over the entire real line with the left value being stretch into a horizontal asymptote at `y=0`.

Answer 2

See explanation

Explanation:

Domain is `RR`

Symmetry
Neither with respect to the `x` axis nor w.r.t the origin.

`arctan(e^(-x))` does not simplify to `arctan(e^x)`
nor to `-arctan(e^x)`

Intercepts
`x` intercepts: none

We cannot get `y = 0` because that would require `e^x = 0`
But `e^x` is never `0`, it only approaches `0` as `xrarr-oo`.
So, `yrarr0` as `xrarr-oo` and the `x` axis os a horizontal
asymptote on the left.

`y` intercept: `pi/4`

When `x=0`, we get `y = arctan(1) = pi/4`

Asymptotes:
Vertical : none

`arctan` is between `-pi/2` and `pi/2` by definition, so never goes to `oo`

Horizontal:
Left: `y=0` as discussed above

Right: `y=pi/2`

We know that, as `thetararrpi/2` with `theta < pi/2`, we get `tantheta rarr oo`
so, as `xrarroo`, we get `e^x rarroo`, so `y=arctan(e^x) rarr pi/2`

First derivative

`y' = e^x/(1+e^(2x))` is never `0` and never undefined, so there are no critical numbers.

For every `x` we have `y' > 0` so the function is increasing on `(-oo,oo)`

There are no local extrema.

Second derivative

`y'' = (e^x(1+e^(2x))-e^x(2e^(2x)))/(1+e^(2x))^2`

`= (e^x+e^(3x)-2e^(3x))/(1+e^(2x))^2`

`=(e^x(1-e^(2x)))/(1+e^(2x))^2`

`y''` is never undefined, and it is `0` at `x=0`

Sign of `y''`:
On `(-oo,0)`, we get `e^(2x) < 1` so `y'' > 0` and the graph is concave up

On `(0,oo)`, we get `e^(2x) > 1` so `y'' < 0` and the graph is concave down

The concavity changes at `x=0`, so the inflection point is:
`(0,pi/4)`

Now sketch the graph