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

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Answer 1 · 2016-12-20T07:19:12

$y in (0, 1]$ . Max y = 1. Points of inflexion : $(+-1, 1/2). larr y = 0 rarr$ is the asymptote.

$y = 1/(1+x^2) >= 1$ and

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

$yin (1, 0]$ . And so,

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

At x =0,

$y'= -(2x)/(1+x^2)^2=0$ , for the turning point (0, 1).

$y'' = -2/(1+x^2)^2+4x^2/(1+x^2)^3$

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

y''' is not 0, when $x = +-1$ .

So, $(+-1, 1/2)$ are the points of inflexion

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

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