If f(x)=x^3-x, what is the average rate of change of f(x) over the interval [1, 5]?

The question is basically asking you to make a straight line between those two points and find its gradient.

The first step youd best take is to find the coordinates of the points needed so:
f(1)=1^3-1 .               f(5)=5^3-5
     =0                              =120
therefore coordinates are:
  (1,0) and (5,120)

the general form of a gradient is:
m=(y2-y1)/(x2-x1)

so plug in your values and you get 
m=(120-0)/(5-1)
   =120/4
   = 30

so your average rate of change between those two values is 30

Hopefully this has been clear enough, good luck

The answer to your question is 1/9