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

May 31, 2017

#### Explanation:

Observe that if $x = \pm a$, $y = {a}^{2} / \left(1 + {a}^{2}\right)$ i.e. for every $\left(x , y\right)$, $\left(- x , y\right)$ too lies on the curve. Hence, the graph is symmetric w.r.t. $y$-axis.

Now as $y = {x}^{2} / \left(1 + {x}^{2}\right) = \frac{1 + {x}^{2} - 1}{1 + {x}^{2}} = 1 - \frac{1}{1 + x} ^ 2 = 1 - {\left(1 + {x}^{2}\right)}^{- 2}$ and hence,

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \left(- \frac{2}{1 + {x}^{2}} ^ 3 \times 2 x\right) = \frac{4 x}{1 + {x}^{2}} ^ 3$

Observe that as $y = 1 - \frac{1}{1 + x} ^ 2$, value of $y$ or range of ${x}^{2} / \left(1 + {x}^{2}\right)$ is limited between $\left[0 , 1\right)$. Further, as $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$ only at $x = 0$ and $x = \pm \infty$ and hence we have an extrema at these points.

Now for second derivative using quotient formula, it is

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{{\left(1 + {x}^{2}\right)}^{3} \times 4 - 4 x \times 3 {\left(1 + {x}^{2}\right)}^{2} \times 2 x}{1 + {x}^{2}} ^ 6$

= $\frac{{\left(1 + {x}^{2}\right)}^{2} \left[4 + 4 {x}^{2} - 24 {x}^{2}\right)}{1 + {x}^{2}} ^ 6 = \frac{4 \left(1 - 5 {x}^{2}\right)}{1 + {x}^{2}} ^ 4$

and this is zero when ${x}^{2} = 0.2$ or $x = \pm 0.447$

Alsso note that when $x = 0$, $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} > 0$, hence wehave a minima at $x = 0$ and when $x \to \pm \infty$, $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} \to 0$, but on the negative side. Further, as $x \to \pm \infty$, $y \to 1$ and hence we have a maxima at $\pm \infty$.

The graph appears as follows:

graph{x^2/(1+x^2) [-5, 5, -1.62, 3.38]}