Let's see.

f(x) = 2x

Make a table:

Code:

___x___ ___f(x)___
0 0 f(0) = 0
1 2 f(1) = 2
2 4 f(2) = 4
3 6 f(3) = 6
4 8 f(4) = 8
5 10 f(5) = 10

So you can find the differences from here,

f(1) - f(0) = 2 - 0 = 2

f(2) - f(1) = 4 - 2 = 2

f(3) - f(2) = 6 - 4 = 2

...

As you see, all differences of f(x)s for consecutive x's are 2. But this is just a basic deduction, it's not proven yet.

a) In a function (for example, f(x) = 3x + 4) you can replace all x's with anything you want.

In your function ( f(x) = 2x ) we will replace all *x*'s with __x+1__ to get a function for f(x+1).

f(x) = 2x

f(x+1) = 2(x+1)

f(x+1) = 2x + 2

Now we have functions f(x) = 2x and f(x+1) = 2x + 2

*"Write a formula that gives the difference between f(x+1) and f(x) in terms of x"*

So we are looking for f(x+1) - f(x)

f(x+1) - f(x) = (2x + 2) - (2x)

f(x+1) - f(x) = 2

Aha, all differences for consecutive *x*'s are 2, we proved it now.