Hi Sue,

The numerator of the difference quotient is s(x + h) - s(x). To evaluate s(x + h) substitute x + h for x in the expression s(x) = x³ + x. Thus

s(x + h) - s(x) = [(x + h)³ + (x + h)] - [x³ + x] = (x + h)³ + (x + h) - x³ - x

To simplify this expression you can first expand (x + h)³ using the binomial theorem but I think it's easier to regroup

s(x + h) - s(x)
= (x + h)³ + (x + h) - x³ - x
= [(x + h)³ - x³] + [ x + h - x]
= [(x + h)³ - x³] + h

and then factor (x + h)³ - x³ as a difference of cubes. Once you have done this and simplified the expression [(x + h)³ - x³] + h you will see that there is a common factor of h. Thus when you form the difference quotient by dividing by h, as long as h is not zero you can cancel the denominator with the common factor of h in the numerator.

Try this and if you have difficulties write back.

Harley

Harley or whomever can still help,

I see the careless mistake I made, however I must not know enough to do the difference of cubes when you have a binomail as one of the cubes. Can you finish out the problem for me Pleeeease! HELP! I can't see where you could cancel the h in the denominator either.

Sue,

The difference of cubes expression is

a³ - b³ = (a - b)(a² + ab + b²)

In your expression a = x + h and b = x so you get

(x + h)³ - x³ = [(x + h) - x][(x + h)² + (x + h)x + x²]

Can you finish it now?

Harley