# How do you sketch the curve f(x)=1/(1+x^2) by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

Dec 20, 2016

$y \in \left(0 , 1\right]$. Max y = 1. Points of inflexion : $\left(\pm 1 , \frac{1}{2}\right) . \leftarrow y = 0 \rightarrow$ is the asymptote.

#### Explanation:

$y = \frac{1}{1 + {x}^{2}} \ge 1$ and

as  x to +-oo, y to 0. S0,

$y \in \left(1 , 0\right]$. And so,

the global maximum is the y-intercept ( x = 0 ) 1.

At x =0,

$y ' = - \frac{2 x}{1 + {x}^{2}} ^ 2 = 0$, for the turning point (0, 1).

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

=2(x^2-1)/(1+x^2)^3=0, when x = +-1. y here = 1/2.

y''' is not 0, when $x = \pm 1$.

So,$\left(\pm 1 , \frac{1}{2}\right)$ are the points of inflexion

graph{y(x^2+1)-1=0 [-2.5, 2.5, -1.25, 1.25]}