# How do you find points of inflection and determine the intervals of concavity given y=tan^-1x?

Feb 21, 2017

See below.

#### Explanation:

$y = {\tan}^{-} 1 x$

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

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

Th denominator is always positive,

so the sign of $y ' '$ is opposite the sign of $x$.

$y ' '$ is positive for $x < 0$ and negative for $x > 0$

The graph is concave up on $\left(- \infty , 0\right)$ and concave down on $\left(0 , \infty\right)$

The inflection point is $\left(0 , 0\right)$.