# Question d6112

Sep 17, 2016

$\left(- 1 , \ln 2\right)$ and $\left(1 , \ln 2\right)$

#### Explanation:

$y = \ln \left(1 + {x}^{2}\right)$

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

y'' = ((2)(1+x^2) - (x^2)(2x))/(1+2x^2)^2 = (2-2x^2)/((1+x^2)^2#

$y ' '$ is defined everywhere and is $0$ at $\pm 1$

Testing the sign of $y ' '$ we find that the sign changes at $- 1$ and at $1$. So the inflection points of $y$ are at $x = - 1$ and $x = 1$.

To finish finding the points, find the corresponding $y$ values.