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Functions [f(x)] and substituton

  1. Aug 22, 2007 #1
    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

    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. jcsd
  3. Aug 22, 2007 #2


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    You should show your work. If f(x)=x^2-2x, then what is f(x+h)?
  4. Aug 22, 2007 #3
    if f(x) = 2x + 1
    then f(x+h) = 2(x+h) + 1
  5. Aug 22, 2007 #4
    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
  6. Aug 22, 2007 #5


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    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.
  7. Aug 22, 2007 #6
    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)?
  8. Aug 22, 2007 #7
    If f(x) = x^2, then f(x+h) = (x+h)^2 = x^2 + 2xh + h^2
  9. Aug 22, 2007 #8
    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!
  10. Aug 22, 2007 #9
    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"
  11. Aug 22, 2007 #10
    better learn this good! you'll encounter it around the ~2nd chapter of your Calculus book.
  12. Aug 22, 2007 #11
    ^^He's got a point [f(x+h) - f(x)]/h will be revisited -- A LOT
  13. Aug 23, 2007 #12
    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?
  14. Aug 23, 2007 #13
    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 :)
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