# What is the domain and range of ln(x^2+1)?

May 31, 2016

Domain is $\mathbb{R}$+, Range is ${\mathbb{R}}^{+}$

#### Explanation:

Domain is given by ${x}^{2} + 1$ >0. That means all real values of x, that is, it would be $\mathbb{R}$

For range, exchange x and y in $y = \ln \left({x}^{2} + 1\right)$ and find the domain. Accordingly, $x = \ln \left({y}^{2} + 1\right)$

${y}^{2} = {e}^{x} - 1$. The domain of this function is all $x \ge 0$ that means all real numbers $\ge$=0

Hence the range of given function would be all Real numbers $\ge 0$