Difference quotient problems.?

Question

Can I get some help with a difference quotient? It seems relatively simple,buti cant quite figure it out.

f(x)=2-(3/x)

((2-(3/(x+h)))-(2-(3/x))/h

Thats really about as far as I get before i get lost. I brought the denominators in the large numerator up as thse respective expressions to the negative first powers, but the sort of complicated things. Dont know if thats the right way to go about it. Any help would be appreciated.

Answer 1

We have:

f(x)=2-(3/x)

Hence:

f(x+h) - f(x) =

((2-(3/(x+h)))-(2-(3/x))/h =

((3/x) - (3/(x+h)))/h =

(3(x+h) - 3x)/(x(x+h)h) =

3h/(x(x+h)h) =

3/(x(x+h))

Answer 2

Remember h represents a small change in x

This is prepaing you for Calculus concepts

The derivative is the limit as delta y with respect to change in x approaches some value like 0 or infinity,etc.

Remember y=F(x) and delta y =F(x+h) - F(x)

Answer 3

and how do you work, without a calculator, 1/3 - 1/7 =, do the same with the numerator, then factor out the "h" and cancel to get the difference quotient....good knowledge of algebra is important in working these difference quotients