Posted by Melanie on Sunday, January 22, 2012 at 12:09am.
f(x)=x/(x^2+x2)
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 7:53am
You 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)(x1). 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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