Question
What is the maximum value of the function y= 18/x^2-9 over the interval -3<x<3?
A. 0
B. 2
C. -2
D. Infinity
A. 0
B. 2
C. -2
D. Infinity
Answers
Damon
do you mean y = 18/(x^2-9) ????
if so
dy/dx = -18 (2x)/(x^2-9)^2
0 at x = 0
In that interval the bigger |x|. the more negative the function so it is max when x = 0
18/-9 = -2
so C
if so
dy/dx = -18 (2x)/(x^2-9)^2
0 at x = 0
In that interval the bigger |x|. the more negative the function so it is max when x = 0
18/-9 = -2
so C