Rolle's theorem

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Image:Rolles theorem 2.png In calculus, Rolle's theorem states that if a function f is continuous on a closed interval [a,b] and differentiable on the open interval (a,b), and f(a) = f(b) then there is some number c in the open interval (a,b) such that

f '(c) = 0.

Intuitively, this means that if a smooth curve is equal at two points then there must be a stationary point somewhere between them. All the assumptions are necessary. For example, if f(x) = |x|, the absolute value of x, then we have that f(-1) = f(1), but there is no x between -1 and 1 for which f '(x) = 0. This is because that function, although continuous, is not differentiable.

The theorem was first stated in India by Bhaskara in 1150, and later by Michel Rolle in 1691.

Rolle's Theorem is used in proving the mean value theorem, which eliminates the requirement that f(a) = f(b).

1 Proof
2 Generalization
3 See also
4 External links

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Proof

The idea of the proof is to argue that if <math>\ f(a) = f(b)</math> then f must attain either a maximum or a minimum somewhere between <math>a</math> and <math>b</math>, and <math>\ f'(x) = 0</math> at either of these points.

Now, by assumption, <math>f</math> is continuous on <math>\ [a,b]</math>, and by the continuity property is bounded and attains both its maximum and its minimum at points of <math>\ [a,b]</math>. If these are both attained at endpoints of <math>\ [a,b]</math> then <math>\ f</math> is constant on <math>\ [a,b]</math> and so <math>\ f'(x) = 0</math> at every point of <math>\ (a,b)</math>.

Suppose then that the maximum is obtained at an interior point <math>\ x \in (a,b)</math> (the argument for the minimum is very similar). We wish to show that <math>\ f'(x) = 0</math>. We shall examine the left-hand and right-hand derivatives separately.

For <math>\ x_1</math> just less than <math>\ x</math>, <math>\ f(x)-f(x_1) \over {x-x_1}</math> is non-negative, since <math>x</math> is a maximum. Thus the limit <math>\ \lim_{x_1 \rightarrow x^-}</math> is non-negative. (Note that we assume that <math>\ f</math> is differentiable to guarantee that the left-hand and right-hand derivatives exist; it does not follow from the other assumptions).

For <math>\ x_2</math> just greater than <math>\ x</math>, <math>\ f(x)-f(x_2) \over {x-x_2}</math> is non-positive. Thus <math>\ \lim_{x_2 \rightarrow x^+}</math> is non-positive.

Finally, since <math>\ f</math> is differentiable at <math>\ x</math>, these two limits must be equal and hence are both 0. This implies that <math>\ f'(x) = 0</math>.

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Generalization

The theorem is usually stated in the form above, but it is actually valid in a slightly more general setting: We only need to assume that f : [a, b] → R is continuous on [a, b], that f(a) = f(b), and that for every x in (a, b) the limit lim_h→0 (f(x + h) − f(x))/h exists or is equal to ±∞.