Expert: Alon Mandes - 5/19/2009

Question
So for this problem:
           6
f(x)=  ------------
      ((x^2)-16)

I got a critical # of zero.
I can't figure out where it's INCREASING/DECREASING, written as intervals.
I also can't figure out where it's CONCAVE UP & CONCAVE DOWN.

I did the deriv test for f'(x) and got it increasing from (-inf,0) but that's wrong and same with decreasing i got (0,inf). for concave up and down by doing the derivative test of f''(x) i got increasing from (-inf, -4), decreasing from (-4,0), increasing from (0,4) and decreasing from (4,inf). Where am I going wrong??

could you help me find where this function is INC/DEC and where its CONCAVE UP/DOWN?? It has to be in interval notation.

Answer
f(x)=6/(x²-16). Let's calculate f'(x) & f''(x) :
f'(x)=-12x/(x²-16)² & f''(x)=48x/(x²-16)³ - 12/(x²-16)² .
To find max/min we solve f'(x)=0 , this gives us x=0. f''(0)<0
that means x=0 is a minimum. We also know that the function f(x)
at x=4 & x=-4 jumps to -∞ ("these are the vertical asymptotes") .
To find increasing/decreasing intervals , we solve f'(x)<0 &
f'(x)<0 . Let's do that :
f'(x)>0 : -12x/(x²-16)²>0 when x<0 Increasing
f'(x)<0 : -12x/(x²-16)²>0 when x>0 Decreasing
To find Concave up/down we solve f''(x)>0 & f''(x)<0 :
f''(x)>0 : 48x/(x²-16)³ - 12/(x²-16)² > 0
          48x/(x²-16)³ > 12/(x²-16)²
          4x/(x²-16) > 1 --> Solution : x>4 & x<-4
f''(x)<0 : 48x/(x²-16)³ - 12/(x²-16)² < 0
          48x/(x²-16)³ < 12/(x²-16)²
          4x/(x²-16) < 1 --> Solution : -4<x<4

Alon.