# Homework Help: Functions [f(x)] and substituton

1. Aug 22, 2007

### Tekee

1. The problem statement, all variables and given/known data
Evaluate and simpliy [f(x+h) - f(x)]/h if f(x) = x^2 -2x

2. Relevant equations
None

3. The attempt at a solution
I just plugged in f(x), or x^2 - 2x for f(X+h) and f(x) and burned it down to h/h, or 1.
That sounds too easy, though...want to know if it's correct.

2. Aug 22, 2007

### cristo

Staff Emeritus
You should show your work. If f(x)=x^2-2x, then what is f(x+h)?

3. Aug 22, 2007

### rocomath

if f(x) = 2x + 1
then f(x+h) = 2(x+h) + 1

4. Aug 22, 2007

### Tekee

Would it simply be x^2 - 2x + h?
(That may be where the problem is...I'm not too sure on that)

If so, I just did:

x^2 - 2x + h - x^2 + 2x --> h ---numerator
h --- denominator

h/h = 1?

EDIT - I see what you did...(rocophysics)
Thanks! I'll try it out.

Last edited: Aug 22, 2007
5. Aug 22, 2007

### cristo

Staff Emeritus
No, it wouldn't. The variable in f(x) is x, whereas the variable in f(x+h) is (x+h): i.e. you need to write (x+h) in place of x in the original function.

6. Aug 22, 2007

### Feldoh

If f(x) = x^2+2x then what would say f(3) be? f(3)=(3)^2+2(3) right? So what is f(x+h)?

7. Aug 22, 2007

### nicktacik

If f(x) = x^2, then f(x+h) = (x+h)^2 = x^2 + 2xh + h^2

8. Aug 22, 2007

### Tekee

Right...so f(x+h) is:

(x+h)^2 - 2x + 2h
I foiled out everything, and I think I'm good to go :)
Thanks for the help!

9. Aug 22, 2007

### Feldoh

f(x) = 2(x)
f(3) = 2(3)
f(x+h) = 2(?)

Other then that (x+h)^2 looks right. :)

All f(x+h) means is go to the function and every where you see an "x" replace it with "x+h"

10. Aug 22, 2007

### rocomath

better learn this good! you'll encounter it around the ~2nd chapter of your Calculus book.

11. Aug 22, 2007

### Feldoh

^^He's got a point [f(x+h) - f(x)]/h will be revisited -- A LOT

12. Aug 23, 2007

### d_leet

But you're original function was x^2-2x so replacing x with x+h you should have
(x+h)^2-2(x+h). Can you expand this? You are off by a sign somewhere in your post, do you see where?

13. Aug 23, 2007

### rocomath

i remember our first test ... lots of ppl did bad b/c of this concept

they sure were happy when they learned how to take the derivative :)