SOLUTION: Look at this function: 𝑓(𝑥) = ln(𝑥^2 + 2𝑥 − 2).
1.) What is the domain?
2.) What are the equations of all asymptotes of 𝑓(𝑥
Algebra.Com
Question 1117852: Look at this function: 𝑓(𝑥) = ln(𝑥^2 + 2𝑥 − 2).
1.) What is the domain?
2.) What are the equations of all asymptotes of 𝑓(𝑥).
3.) Find the solution set (solve) for the inequality 𝑓(𝑥) ≤ 0, write the solution set in interval notation.
Found 2 solutions by solver91311, Boreal:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
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The domain of the function is the set of all real numbers excluding all values of the independent variable such that the argument of the natural logarithm function is less than or equal to zero.
\ =\ \ln\(x^2\ +\ 2x\ -\ 2})
\ =\ \{x\,\in\,\mathbb{R}\,|\,x^2\ +\ 2x\ -\ 2\ >\ 0\})
Vertical asymptotes are located where the independent variable equals a zero of the logarithm argument.
The solution set for the inequality \ \leq\ 0)
is the set of all values of the independent variable such that the argument of the logarithm function is in the interval (0,1].
(x\ +\ 3))
So the solution set for is:
John
My calculator said it, I believe it, that settles it
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Look first at the graph of 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
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