# How do you sketch the graph y=(2e^x)/(1+e^(2x)) using the first and second derivatives?

Aug 8, 2017

First we note that as ${e}^{2 x} > 0$ the denominator of the quotient never vanishes, so that:

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

is defined and continuous for all $x \in \mathbb{R}$.

By writing the function as:

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

we can also see that the function is even and at the limits of the domain of definition we have:

${\lim}_{x \to - \infty} \frac{2}{{e}^{-} x + {e}^{x}} = {\lim}_{x \to \infty} \frac{2}{{e}^{-} x + {e}^{x}} = 0$

So that $f \left(x\right)$ has $y = 0$ as an horizontal asymptote on both sides.

Evaluate now the first and second derivatives:

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\frac{1}{\cosh}\right) = - \sinh \frac{x}{\cosh} ^ 2 x$

$\frac{{d}^{2} f}{\mathrm{dx}} ^ 2 = - \frac{d}{\mathrm{dx}} \left(\sinh \frac{x}{\cosh} ^ 2 x\right) = - \frac{{\cosh}^{3} x - 2 {\sinh}^{2} x \cosh x}{\cosh} ^ 4 x = \frac{2 {\sinh}^{2} x - {\cosh}^{2} x}{\cosh} ^ 3 x = \frac{{\sinh}^{2} x - 1}{\cosh} ^ 3 x$

We have then that $\frac{\mathrm{df}}{\mathrm{dx}} > 0$ for $x < 0$ and $\frac{\mathrm{df}}{\mathrm{dx}} < 0$ for $x > 0$ while the only stationary point where $\frac{\mathrm{df}}{\mathrm{dx}} = 0$ is for $x = 0$.

The function is therefore strictly increasing in $\left(- \infty , 0\right)$, strictly decreasing in $\left(0 , + \infty\right)$ and reaches a maximum for $x = 0$ ov value $f \left(0\right) = 1$. As $\cosh x \ge 1$ we can see that this is an absolute maximum.

The second derivative is null for $\sinh x = \pm 1$, that is for:

$x = \ln \left(\sqrt{2} \pm 1\right)$

These are two inflection points and the function is concave down in the interval $\left(\ln \left(\sqrt{2} - 1\right) , \ln \left(\sqrt{2} + 1\right)\right)$and concave up outside this interval.

graph{ (2e^x)/(1+e^(2x)) [-3.59, 3.59, -1.794, 1.796]}