Difference Quotient Problem.

A carpenter is constructing a large cubical storage shed. The volume of the shed is given by V(x) = x^2[(16-x^2)/4x], where x is the side length, in meteres.

a) Simplify the expression for the volume of the shed.

**I had no clue how to simplify it, but the answer is V(x)=4x-1/4x^3**

b) Determine the average rate of change of the volume of the shed when the side lengths are between 1.5m and 3m.

**Am I supposed to use the difference quotient to find average rate of change? I assumed so, so this is my work which ended up being wrong:**

V(x)=4x-1/4x^3

delta y/delta x= [f(x+h)-f(x)]/h

=[[4(x+h)-1/4(x+h)^3]-[4(x)-1/4(x)^3]]/h *is that what i was supposed to do..? so skipping a few steps I got:

=-1/4x^3 - 3/4x^2h - 3/4xh^2 - 1/4h^3 +4x +4h *skipping a few more steps:

=-3/4x^2 - 3/4xh - 1/4h^2 + 4

So I know to sub in 1.5 for h, since 1.5m is the interval between the two points (1.5 and 3). But what would I sub in for x? I chose to sub in 1.5 for it also.. can someone explain why if thats right?

So when I subbed those numbers in I get -7.75.. the answer is 0.0625m^3/m

c) Determine the instantaneous rate of change of the colume of the shed when the side length is 3m.

**Would I just sub in 3 for x and 0.000001 (a number really close to 0) for h, since the instantaneous interval is 0? Answer is -2.75m^3/m**

I'm having a bit of trouble doing difference quotient stuff =/