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July 31, 2014

July 31, 2014

Posted by **Melanie** on Sunday, January 22, 2012 at 12:09am.

find:

x intercepts

vertical asymptotes

horizontal/slant asymptotes

derivative

second derivative

increasing interval

decreasing interval

concave up interval

concave down interval

- math -
**drwls**, Sunday, January 22, 2012 at 7:53amYou should be able to do some of these yourself. If not, you are behind class level and would benefit from private tutoring.

The x intercept(s) occur where f(x) = 0. That can only happen at when the numerator is zero. In other words, at x = 0.

Vertical asymptotes occur wherever the denominator of f(x) is zero. Note that the denominator of f(x) can be written (x+2)(x-1). That should tell you where the asymptotes are.

For the derivative of

f(x) = x/(x^2 +x -2), let

u(x) = x and v(x) = x^2 +x -2

and use the rule for the derivative of the fraction u(x)/v(x), which you should know.

f'(x) = (v*u' - u*v')/v^2

= [x^2 +x -2 -x*(2x +1)]/(x^2 +x -2)^2

= (-x^2 -2)/(x^2 +x -2)^2

Use the same "u/v" rule for the second derivative.

Plotting the f(x) function yourself should help for the last four questions. "Increasing up or down" is determined by the sign of f'(x).

"Concave up or down" is determined by the sign of f''(x), the second derivative.

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