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Mean value theorem question

  1. May 1, 2005 #1
    so by the defn:

    suppose that f is continuous on a closed I:= [a,b] and that f has a derivative in the open interval (a,b). then tehre exists at least one point c in (a,b) st f(b) - f(a) = f'(c)(b - a).

    ok, so what if I put this in terms of f'(c)? isn't that the definition of the derivative?

    So it's saying that if f has a derivative in (a,b), then there is a point c that has a derivative?

    I"m kind of lost....because this sounds a bit redundant.

    I"m kind of having trouble on what the mean value theorem is telling me....
  2. jcsd
  3. May 1, 2005 #2


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    Yes that's another way to define the derivative. The Theorem is saying that if f(x) has a derivative in (a,b), then there is a number c in (a,b) such that is derivative is equal to what you saw.

    If you look up Rolle Theorem you might understand more the importance of the mean value theorem.
  4. May 1, 2005 #3
    Starting with the equation:

    f(b) - f(a) = f'(c)(b - a)

    and dividing by (b - a) gives

    f'(c) = [f(b)-f(a)]/(b - a)

    Can you see what that fraction on the right hand side is?
  5. May 2, 2005 #4


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    No, that's NOT the definition of the derivative: (f(b)-f(a))/(b-a) is the slope of the straight line from (a,f(a)) to (b,f(b)) a secant line rather than a tangent line. The derivative requires a limit.

    Geometrically what the mean value theorem says is that if f(x) is differentiable between a and b and continuous (from the left and right) at a and b, then there exist some point between a and b where the tangent line is parallel to the the straight line between a and b. The reason for the name "mean value" theorem is that (f(b)-f(a))/(b-a) is a sort of "average" ("mean") value of the derivative at all points between a and b- and the mean value theorem says that there is some point where the actual derivative is the same as the average derivative.

    The mean value theorem is crucial in dealing with "anti-derivatives". You know that many functions have the same derivative: f(x)= 4, g(x)= 1000 both have the same derivative: 0; f(x)= x2+ 10 and g(x)= x2+ 100000 both have the same derivative: 2x. Of course, they differ by a constant and the derivative of a constant is 0 (from the basic definition). But how do we know there are not other, perhaps extremely complicated functions that also have that derivative? How do we know that there are not other, perhaps extremely complicated functions that have derivative 0 for all x? From the mean-value theorem. If f'(x)= 0 for all x, then for any a, b,
    f(b)-f(a)= f'(c)(b-a)= 0(b-a)= 0. In other words, f(a)= f(b) for any a and b: f is a constant.

    As a reminder that f(x) continuous at the endpoints is necessary, consider
    f(x)= 2x if x< 3, f(x)= 100 if x>= 3. f is differentiable between 0 and 3 and its derivative there is 2. If f(3)- f(0)= 2(3-0)?
  6. May 2, 2005 #5


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    Drawing a picture may help. If you draw a straight line from (a,f(a)) to (b,f(b)), then the mean value theorem says that there is a point in (a,b) such that its tangent is parallel to that line.
    In other words: There is some point at which the rate of change equals the average change.
    Intuitively this is clear. It says for example that if a car has travelled 180 km in 2 hours, there must have been a point where it went 90 km/h.

    Or, if two runners have a race and start and finish at the same time, then there must've been a point where their speed where equal.
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