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Mean Value Theorem

  1. Oct 10, 2004 #1
    If f is continous in [a,b] and differentiable in (a,b), and if xE(a,b), then there exists u in (a,b) such that
    f(x) = f(a)+(x-a)f'(u)

    What's wrong if i state :
    (i)If f is continous in [a,b] and differentiable in [a,b], and if xE[a,b]
    (ii)If f is continous in (a,b) and differentiable in (a,b), and if xE(a,b)

  2. jcsd
  3. Oct 10, 2004 #2
    usually we write,
    f'(u) = (f(x)-f(a))/(x-a) .. (1)

    here x can take value of a ..
    see what happens to (1) when x takes value of a?

    it is easy to note that if f is continuous in (a,b) then it is continuous in [a,b]
    cuz f(a) = lim_{x->a+}f(x)
    (we don't bother about x->a-)
    f(b) = lim_{x->b-}f(x)
    (we don't bother about x->b+)

    -- AI
  4. Oct 10, 2004 #3
    In the second case, f may tend to infinity at a or at b. So f being continuous on [a, b] is required.
  5. Oct 10, 2004 #4
    f(a)=lim{x->a} f(x) is the definition of continuity at a, so this is just assuming what you are trying to show.

    If you don't assume continuity at a then f(a) can have whatever value you like, unrelated to the rest of f(x). e.g.
    f(0)=1, f(1)=1,f(x)=0 for x in (0,1).
  6. Oct 10, 2004 #5
    in the second case,
    f(x) is defined over (a,b) which excludes a and b
    so f(a) and f(b) are actually not defined originally and hence cannot be infinity
    so i can define continuity extending to both extremities and still the theorem would be true

    [Edit]eeps! Wong is right on one thing here
    say the function is 1/x and the interval is (0,1) the limx->0+ f(x) would be infinity .... So i suppose we could add a condition that lim x->a+ f(x) and limx->b-f(x) must exist and must be finite..... then it is extendible as i said earlier[/Edit]

    definition of continuity requires that,
    f(a) = limx->a- f(x) = limx->a+ f(x)
    (this is a strict necessity)
    But if u see what i said in my earlier post, when we talk of an interval like (a,b) i can skip one of the conditions since my viewport now is restricted to interval (a,b) alone.

    -- AI
    Last edited: Oct 10, 2004
  7. Oct 10, 2004 #6


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    In other words, you're assuming f is a function continuous on [a, b]. :tongue2:
  8. Oct 10, 2004 #7


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    There is nothing WRONG with assuming f is differentiable on [a,b] except that it is not as general. The theorem is true even if f is NOT differentiable at a or b.
    (Consider the function f(x)= |x| and apply the mean value theorem on [0, 1].)

    On the other hand you MUST have f continuous at a and b since the values used in the theorem are calculated there.

    For example, if f(x)= x for x NOT equal to 0 or 1 but f(0)= -1, f(1)= 2. Can you find c between 0 and 1 such that f'(c)= f(1)- f(0) ?
  9. Oct 11, 2004 #8
    Well i did edit my post indicating that fact :biggrin:
    Though i am not exactly assuming it :tongue2:
  10. Oct 11, 2004 #9


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    If I got this right, you can't.

    This is because it is not continuous at 0 and 1.

    Continuity does not imply differentiability.

    When they say it is continuous on the interval, they are saying:

    f(a)=the limit, as x approaches a, is equal to f(a). Again, it has been mentionned earlier that the right-hand limit must equal the left-hand limit. That is continuity. It may be continuous on the left or right, but the theorem has mentionned nothing about it, so it shall be excluded in this case.

    Feel free to write your own theorem.
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