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

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Answer 1 · 2017-05-03T13:02:06

See graph and analysis below.

$y=ln(x^2+1)$

Since $x^2 +1 >= 1 forall x in RR ->y$ is defined $forall x in RR$

$:.$ the domain of $y$ is $(-oo,+oo)$

Since $(x^2+1) >= 1 forall x in RR$ then $ln(x^2+1) >=0 forall x in RR$

$:.$ the range of $y$ is: $[0, +oo)$

$y=0 -> ln(x^2+1) = 0$

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

$x^2=0 -> x=0$

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

The graph of $y$ is shown below.

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

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