Question

Find the extrema of \(\dfrac{x^2 + x - 6}{x^2 - 1}\)

Answer 1

http://www.wolframalpha.com/input/?i=factor+x%5E3-15+x%5E2%2B3+x-5

Answer 2

I presume you equated it to 0. If so, try using the Factor Theorem.

Answer 3

Technically, i am trying to find the extrema of x^2 +x-6 / (x^2-1), I have the first derivative and its critical numbers, i know what my max and min point is. but i am stuck in trying to make the second derivavtive test. the second deriv is 2(x^3-15x^2+3x-5/(x^2-1)^3. F"(x)= DNE never because the domain is all real numbers but F"(x)=0 is killing me. \[x ^{3}-15x ^{2}+3x-5 =0\] I have tried everything.... i think..

Answer 4

And why would you set f''(x)=0? f'(x) = 0 for finding stationary points, and we then substitute those x-values into f''(x) to determine if they are max or min pts.

Answer 5

I dont know my professor taught us that the second deriv test was setting to zero, and finding any #'s that make it DNE and set them on a numberline and test the intervals for + or - signs and then according to the sign determining the concavity

Answer 6

@orezsnel : Alright, after considering your question for some time, I get where you are coming from (I believe it might be due to some miscommunication) - setting f''(x)=0 ensures uniformity in f'(x) and if f'(x)=0, this would represent a point of inflexion, while for f''(x) to be inadmissible allows for the localization of potential vertical asymptotes, hence we use the range defined by the bounds set between these values of f''x to determine the signs of f'(x) for determining the ascent/descent of the curve. So returning to your initial question, you have found f''(x) & using the link provided by @eliassaab initial post, we can see that this is zero when x=14.8203 (i.e. this is a point of inflexion of the original curve) while x=1 is a vertical asymptote (since it causes f''(x) to be inadmissible) so you would now test f''(x) for values above 14.8203 (e.g. x=15) as well as 1<x<14.8203 (e.g. x=4) to determine the signs of f'(x) and thus elucidate the concavity/shape of the curve near the point of inflexion (where x=14.8203).

Answer 7

@orezsnel : An alternative & perhaps simpler method in this instance [since f'(x) looks less complicated than f''(x)] would also be to simply test f'(x) for values above 14.8203 (e.g. x=15) as well as 1<x<14.8203 (e.g. x=4) to determine the signs of f'(x).