f(x)= x^3/x^2-16 defined on [-19, 16]
How would you find the vertical asymptopes? and the inflection point?
I think you take the derivative and set it to zero? but i cant figure out the correct derivative or something??
calc - Damon, Wednesday, March 18, 2009 at 3:51pm
assume you mean (x^2-16) on the bottom
vertical asymptotes are where the denominator goes to zero
that is when x = -4 and x = +4
derivative of fraction =
(bottom * derivative of top - top * derivative of bottom)
all over bottom squared
[(x^2-16)*3 x^2 - x^3(2x) ]/bottom^2
look for zeros of numerator
3 x^4 - 48 x^2 - 2 x^4
=x^4 - 48 x^2
= x^2 (x^2-48)
= x^2 (x-sqrt 48)(x+sqrt 48)
so zero at -sqrt 48, 0 and + sqrt 48
check my arithmetic, I did that fast.
calc - Damon, Wednesday, March 18, 2009 at 3:52pm
by the way sqrt 48 = 4 sqrt 3