# Question #92256

##### 2 Answers

See explanation

#### Explanation:

Break this into two parts, firstly the inner part:

This is positive and increasing for all real numbers and goes from 0 to

The we have:

The has a right horizontal asymptote at

These points therefore get pulled to

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

Which is the positive part of the

See explanation

#### Explanation:

**Domain** is

**Symmetry**

Neither with respect to the

#arctan(e^(-x))# does not simplify to#arctan(e^x)#

nor to#-arctan(e^x)#

**Intercepts**

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.

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:

Right:

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**

For every

There are no local extrema.

**Second derivative**

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

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

Sign of

On

On

The concavity changes at

Now sketch the graph