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]}