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

1 Answer
Jun 15, 2016

Answer:

Please see below.

Explanation:

As the domain for #lnx# is #x>0#, the domain for #ln(1+x)# is #x> -1#. However range is #-oo < ln(1+x) < oo#.

Function is continuous and relation is one to one.

Again for #0 < x < 1#, #lnx# is negative and hence #ln(1+x)# is negative for #-1 < x < 0#.

And as #x->-1#, #ln(1+x)->-oo# and hence #x+1=0# is asymptote for #ln(1+x)#.

At #x=0#, #ln(1+x)=0# (it also means #ln(1+x)# cuts #x#-axis at #(0.0)# and for #x>0# #ln(1+x)>0#.

One can also take additional values for #x# such as #{1,2,3,4,5,6,7,8,9,10}# for which #ln(1+x)# would be #{0.693,1.099,1.386,1.609,1.792,1.946,2.079,2.197,2.303,2.398,2.485}#. The graph of #ln(1+x)# appears as given below.

graph{ln(1+x) [-5.415, 14.585, -6, 4]}