Question

find the critical points of (4x)/(x^2-9)

Answer 1

Critical points at x=0,3, -3.

Answer 2

tried those already, it keeps saying their wrong. :l

Answer 3

http://gyazo.com/bfb06971563e78a95d23a832eec58560

Answer 4

That's strange. You may have to list them from least to greatest, or greatest to lease.

Answer 5

you need to take the derivative first, then find the critical points

Answer 6

did that. still cant seem to get the right answer

Answer 7

i take first deriv. set numerator equal to zero. solve for x
i get -2.16, and 4.16

Answer 8

derivative has no zeros
it is always negative function is always decreasing

Answer 9

i use the quadratic formula

Answer 10

there are no critical points, unless you want to count \(-3\) and \(3\) which should not count because they are not in the domain of the function, and therefore are not in the domain of the derivative

Answer 11

not even 0 is a critcal point?

Answer 12

derivative is
\[\frac{-(4x^2+36)}{(x^2-9)^2}\]

Answer 13

numerators is never zero, so no 0 is not a critical point

Answer 14

i guess you can say 3 and -3 are if you have to say something, but as i said they are not in the domain of the function to begin with

Answer 15

@satellite73 Would you mind explaining to me why -2.16 and 4.16 cant be the critical points? I've been looking around online, and every thing says that where every the derivative = 0, is a critical point, unless that point is not in the domain. 3 and -3 arent in the domain, and i get that, but -2.16 and 4.16 are.