# How do you sketch the graph y=1/(1+x^4) using the first and second derivatives?

Jan 10, 2017

$y \left(x\right)$ is positive and even, infinitesimal for $x \to \pm \infty$, with a maximum in $x = 0$ and two inflection points for $x = \pm \sqrt{\frac{3}{5}}$

#### Explanation:

To analyze the behavior of the function we start from considering the domain of definition and see that $y \left(x\right)$ is defined and continuous in all of $\mathbb{R}$ as the denominator never vanishes.

We can also see that $y \left(x\right)$ is positive in its domain and it it even as

$y \left(- x\right) = y \left(x\right)$

so that its graph is symmetrical with respect to the $y$-axis.

At the limits of the domain $y \left(x\right)$ is infinitesimal as:

${\lim}_{x \to \pm \infty} y \left(x\right) = 0$

so the function is going to have $y = 0$ as asymptote on both sides.

Now we calculate the first and second derivatives:

$y ' \left(x\right) = - \frac{4 {x}^{3}}{1 + {x}^{4}} ^ 2$

$y ' ' \left(x\right) = \frac{- 12 {x}^{2} {\left(1 + {x}^{4}\right)}^{2} + 32 {x}^{6} \left(1 + {x}^{4}\right)}{{\left(1 + {x}^{4}\right)}^{4}} = \frac{- 12 {x}^{2} \left(1 + {x}^{4}\right) + 32 {x}^{6}}{{\left(1 + {x}^{4}\right)}^{3}} = \frac{4 {x}^{2} \left(- 3 {x}^{4} + 8 {x}^{4} - 3\right)}{{\left(1 + {x}^{2}\right)}^{3}} = \frac{4 {x}^{2} \left(5 {x}^{4} - 3\right)}{{\left(1 + {x}^{2}\right)}^{3}}$

We can see that the only critical point where $y ' \left(x\right) = 0$ is for $x = 0$ and that:

$x < 0 \implies y ' \left(x\right) > 0$
$x > 0 \implies y ' \left(x\right) < 0$

so that the function is monotone increasing for $x \in \left(- \infty , 0\right)$, reaches a maximum for $x = 0$ and then is monotone decreasing for $x \in \left(0 , + \infty\right)$.

The second derivative has three zeros in $x = 0$ and $x = \pm \sqrt{\frac{3}{5}}$. We know the first is a maximum, so $y \left(x\right)$ has two inflection points in $x = \pm \sqrt{\frac{3}{5}}$. and:

$\left\mid x \right\mid > \sqrt{\frac{3}{5}} \implies y ' ' \left(x\right) > 0$ and y(x) is concave up.
$\left\mid x \right\mid < \sqrt{\frac{3}{5}} \implies y ' ' \left(x\right) < 0$ and y(x) is concave down.

graph{1/(1+x^4) [-10, 10, -5, 5]}