# Question #92256

Aug 4, 2015

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 $\infty$ as $x$ goes from $- \infty$ to $\infty$

The we have:

$\arctan \left(u\right)$

The has a right horizontal asymptote at $y = \frac{\pi}{2}$. Going from $u = 0 \rightarrow \infty$, at $u = 0$ this function is positive and increasing over this domain, takes a value of 0 at $u = 0$, a value of $\frac{\pi}{4}$ at $u = 1$ and a value of $\frac{\pi}{2}$ at $u = \infty$.

These points therefore get pulled to $x = - \infty , 0 , \infty$ 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 $y = 0$.

Aug 4, 2015

See explanation

#### Explanation:

Domain is $\mathbb{R}$

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

$\arctan \left({e}^{- x}\right)$ does not simplify to $\arctan \left({e}^{x}\right)$
nor to $- \arctan \left({e}^{x}\right)$

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 $x \rightarrow - \infty$.
So, $y \rightarrow 0$ as $x \rightarrow - \infty$ and the $x$ axis os a horizontal
asymptote on the left.

$y$ intercept: $\frac{\pi}{4}$

When $x = 0$, we get $y = \arctan \left(1\right) = \frac{\pi}{4}$

Asymptotes:
Vertical : none

$\arctan$ is between $- \frac{\pi}{2}$ and $\frac{\pi}{2}$ by definition, so never goes to $\infty$

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

Right: $y = \frac{\pi}{2}$

We know that, as $\theta \rightarrow \frac{\pi}{2}$ with $\theta < \frac{\pi}{2}$, we get $\tan \theta \rightarrow \infty$
so, as $x \rightarrow \infty$, we get ${e}^{x} \rightarrow \infty$, so $y = \arctan \left({e}^{x}\right) \rightarrow \frac{\pi}{2}$

First derivative

$y ' = {e}^{x} / \left(1 + {e}^{2 x}\right)$ 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 $\left(- \infty , \infty\right)$

There are no local extrema.

Second derivative

$y ' ' = \frac{{e}^{x} \left(1 + {e}^{2 x}\right) - {e}^{x} \left(2 {e}^{2 x}\right)}{1 + {e}^{2 x}} ^ 2$

$= \frac{{e}^{x} + {e}^{3 x} - 2 {e}^{3 x}}{1 + {e}^{2 x}} ^ 2$

$= \frac{{e}^{x} \left(1 - {e}^{2 x}\right)}{1 + {e}^{2 x}} ^ 2$

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

Sign of $y ' '$:
On $\left(- \infty , 0\right)$, we get ${e}^{2 x} < 1$ so $y ' ' > 0$ and the graph is concave up

On $\left(0 , \infty\right)$, we get ${e}^{2 x} > 1$ so $y ' ' < 0$ and the graph is concave down

The concavity changes at $x = 0$, so the inflection point is:
$\left(0 , \frac{\pi}{4}\right)$

Now sketch the graph