# How do you graph y=ln(x^2 +1)?

May 3, 2017

See graph and analysis below.

#### Explanation:

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

Since ${x}^{2} + 1 \ge 1 \forall x \in \mathbb{R} \to y$ is defined $\forall x \in \mathbb{R}$

$\therefore$ the domain of $y$ is $\left(- \infty , + \infty\right)$

Since $\left({x}^{2} + 1\right) \ge 1 \forall x \in \mathbb{R}$ then $\ln \left({x}^{2} + 1\right) \ge 0 \forall x \in \mathbb{R}$

$\therefore$ the range of $y$ is: $\left[0 , + \infty\right)$

$y = 0 \to \ln \left({x}^{2} + 1\right) = 0$

Then: ${x}^{2} + 1 = {e}^{0} = 1$

${x}^{2} = 0 \to x = 0$

Hence: $y = 0$ at $x = 0$

The graph of $y$ is shown below.

graph{ln(x^2+1) [-10, 10, -5, 5]}