Thread: Finding the Range using f '(x)

I need to find the range of f(x)=(x^2-8)/(x+3) and use f'(x) to justify my answer. I found f'(x) to be (x^2+6x+8)/(x+3)^2 and I found the critical numbers of f to be -4,-3,-2, but I don't know where to go from here. By looking at the graph the range looks like its (-inf,-8)(-4,+inf) but I don't know how to prove that algebraically. Please Help!

For your derivative numerator, in f'(x), your two critical points tell where the tangent slope is zero.

Evaluating f(x) for these, the greater value of f(x) is above the other,
you find that if the local maximum is below the local minimum,
then in between these is a range of values of y that are not on the curve.

The critical point x= -3 gives the value of x not belonging to the domain.