Question 1117852
Look first at the graph of x^2+2x-2
{{{graph(300,300,-10,10,-10,10,x^2+2x-2)}}}

The roots are done using the quadratic formula
x=(1/2){-2+/- sqrt(4+8)}
This is -1+/-sqrt(3) or -1+sqrt(3) or x > 0.732.
This is -1-sqrt(3) or x < -1.732
When x is between those values, the ln does not exist
The domain is therefore (-oo, -2.732) and (0.732, oo)
From the left side, as x approaches -2.732 from the negative, or to 0.732 from the right, the ln becomes closer to 0 and the ln becomes large negative.
The asymptotes are x=-1-sqrt(3) and x= -1+sqrt (3)
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For the function being <0, one needs values of the quadratic less than 1, since ln (1) is 0, and greater than 0, since ln 0 doesn't exist.
x^2+2x-2=1
x^2+2x-3=0
(x+3)(x-1)=0
x=1, -3
So (-3, -1-sqrt(3)) and (-1+sqrt(3), 1) the function is < 0



{{{graph(300,300,-5,5,-10,10,ln(x^2+2x-2))}}}