Finding Critical Numbers and Absolute Extrema's.

1) $\displaystyle {\frac{3}{2}}x^{\frac{1}{2}}-{\frac{3}{2}}x^{\frac{-1}{2}}$

For this one... I found that my critical number was 1. I don't know if its right.

I was also give the Interval [-2,3]. So I plugged it all into my original function of $\displaystyle f(x)=(x-3)x^{\frac{1}{2}}$

After I plugged it in, I got

f(1)=2

f(3)=0

f(-2)=undefined.

But then again, I don't know if 1 is the critical number. And.. since it's undefined.. I just leave it alone right? So the max would be (1,2) and min would be (3,0) assuming that 1 is the critical number?

2) $\displaystyle g(x) = \frac{-4x+6}{(x^2-3x)^3}$

Found that the critical number was $\displaystyle {\frac{3}{2}}$

Given interval [-1,6]

I need help finding the absolute extremas here even though its just plugging in (Lipssealed)

Suppose to plug it into this original function:

$\displaystyle g(x)=(x^2-3x)^{-2} + 2$