Math CentralQuandaries & Queries


Question from Sue, a student:

When s(x)=x^3+x, compute and simplify the difference quotient s(x+h)-s(x)/h.

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 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.


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?


About Math Central


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
Quandaries & Queries page Home page University of Regina PIMS