Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Mean theorim question

  1. Nov 21, 2008 #1
    i know that h(x)=f(x)-g(x)
    but on what basis they conclude that

    h'(c)=f'(c)-g'(c)


    http://img356.imageshack.us/img356/5427/69700925ii7.gif [Broken]

    ??
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 21, 2008 #2
    If h(x) = f(x) for all x then h'(x) = f'(x).
     
    Last edited: Nov 21, 2008
  4. Nov 21, 2008 #3
    for a fact
    ??
     
  5. Nov 21, 2008 #4
    Yes; consider the definition [tex]h'(x) = \lim_{t\to 0}\frac{h(x+t)-h(x)}{t}[/tex].
     
  6. Nov 21, 2008 #5
    If f(x)=g(x) for all x, then the two functions are the same. And for this very reason they have the same derivative.
     
  7. Nov 21, 2008 #6
    but f(x) is a curve and g(x) is a line
    f(x) is not g(x)
    and so is h(x) differs them all
    look at the link

    http://img356.imageshack.us/img356/5427/69700925ii7.gif [Broken]

    all the lines are different

    h(c)=f(x)-g(x)

    how they came to
    h'(c)=f'(x)-g'(x)
     
    Last edited by a moderator: May 3, 2017
  8. Nov 21, 2008 #7
    Now you're just confused by notation. Let's use some neutral letters.

    If A and B are differentiable functions on [a,b] such that A(x) = B(x) for all x in [a,b], then A'(x) = B'(x).

    Why? [tex]A'(x) = \lim_{t\to 0}\frac{A(x+t)-A(x)}{t} = \lim_{t\to 0}\frac{B(x+t)-B(x)}{t} = B'(x)[/tex].

    Also, they said "h'(c)=f'(c)-g'(c)" NOT "h'(c)=f'(x)-g'(x)".
     
  9. Nov 21, 2008 #8
    when you say "A(x) = B(x) for all x in [a,b]"

    that mean that they share each one of their points
    which means that A(x) and B(x) are the same function then of course A'(x) = B'(x) (its the same function)

    which is not true in the graph
     
    Last edited: Nov 21, 2008
  10. Nov 21, 2008 #9
    Take A(x)=h(x) and B(x)=f(x)-g(x)
     
  11. Nov 21, 2008 #10
    ahhhh thanks
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook