Question 1117852
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*[illustration ln(x_2_2x_2).jpg]


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.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \ln\(x^2\ +\ 2x\ -\ 2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \text{dom}\(f\)\ =\ \{x\,\in\,\mathbb{R}\,|\,x^2\ +\ 2x\ -\ 2\ >\ 0\}]


Vertical asymptotes are located where the independent variable equals a zero of the logarithm argument.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ -\ 2\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -1\ -\ \sqrt{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -1\ +\ \sqrt{3}]


The solution set for the inequality *[tex \Large f(x)\ \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].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ -\ 2\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 1)(x\ +\ 3)]


So the solution set for *[tex \Large f(x)\ \leq\ 0] is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \{x\,\in\,\mathbb{R}\,|\,-3\ \leq\ x\ <\ -1\ -\sqrt{3}\ \small{\vee}\LARGE\ -1\ +\ \sqrt{3}\ <\ x\ \leq\ 1\}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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