Question

what is the absolute maximum and absolute minimum of g(x)= 1/x^2-9; on the specific interval 0 is less than or equal to x, and x is less than or equal to 2

Answer 1

so you are looking at

\[g(x) = \frac{1}{x^2 - 9}\]

between \[0\le x \le2\]

Answer 2

yes

Answer 3

ok... so the graph has asymptotes y = 0, x = -3, 3

and looks like



so for me, knowing what the graph looks like, I'd substitute x = 0 and x = 1 into the equation...

so abs max(-1/9) and bs min -1/5