SOLUTION: Using the functions f(x)=(x^2-1)/(x^2-x-2)and g(x)=(x^3-1)/(x^2-1) identify the domain, vertical asymptote, holes, horizontal asymptote or slant asymptote, and the Left and Right E
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Question 674273: Using the functions f(x)=(x^2-1)/(x^2-x-2)and g(x)=(x^3-1)/(x^2-1) identify the domain, vertical asymptote, holes, horizontal asymptote or slant asymptote, and the Left and Right End Behaviors.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Factor the numerator and denominator polynomials:
\ =\ \frac{x^2\ -\ 1}{x^2\ -\ x\ -\ 2}\ =\ \frac{(x\ -\ 1))x\ +\ 1)}{(x\ -\ 2)(x\ +\ 1))
Values that make the denominator equal to zero: 2, -1, so:
Since the factor 
only occurs in the denominator, there is a vertical asymptote at
Since the factor 
occurs in both the numerator and the denominator, there is a hole at:
\right))
\ =\ \frac{\infty}{\infty})
which is a L'Hôpital determinate form, so take the first derivatives of the numerator and denominator functions:
\ =\ \lim_{x\rightarrow-1}\,\frac{\frac{d}{dx}(x^2\,-\,1)}{\frac{d}{dx}(x^2\,-\,x\,-\,2)}\ =\ \lim_{x\rightarrow-1}\,\frac{2x}{2x\ -\ 1}\ =\ \frac{2}{3})
So the hole is at:
)
Rational functions where the degree of the numerator polynomial is equal to the degree of the denominator polynomial have a horizontal asymptote at 
where 
is the lead coefficient of the numerator polynomial and 
is the lead coefficient of the denominator polynomial. Since this rational function has a horizontal asymptote, it cannot have a slant asymptote.
Left behavior:
\ =\ \lim_{x\rightarrow-\infty}\,\frac{x^2\,-\,1}{x^2\,-\,x\,-\,2}\ =\ frac{\infty}{\infty})
Again, we have the L'Hôpital determinate form, so take the first derivatives of the numerator and denominator functions:
\ =\ \lim_{x\rightarrow-\infty}\,\frac{\frac{d}{dx}(x^2\,-\,1)}{\frac{d}{dx}(x^2\,-\,x\,-\,2)}\ =\ \lim_{x\rightarrow-\infty}\,\frac{2x}{2x\ -\ 1}\ =\ \frac{\infty}{\infty})
We still have the determinate form, so take the 2nd deriviative:
\ =\ \lim_{x\rightarrow-\infty}\,\frac{\frac{d^2}{dx^2}(x^2\,-\,1)}{\frac{d^2}{dx^2}(x^2\,-\,x\,-\,2)}\ =\ \lim_{x\rightarrow-\infty}\,\frac{2}{2}\ =\ 1)
Hence the function tends to the value 1 as the independent variable decreases without bound.
Similar analysis, left as an exercise for the student, is sufficient to describe the Right behavior.
John
My calculator said it, I believe it, that settles it
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