In calculus, Rolle's theorem states that if a function f is continuous on a closed interval ^* 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 at x=0.

The statement of the theorem was first elucidated by Indian astronomer Bhaskara in the 12th century. The theorem was reproduced centuries 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).

Proof

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The idea of the proof is to argue that if $\backslash \; f(a)\; =\; f(b)$ then f must attain either a maximum or a minimum somewhere between $a$ and $b$, and $\backslash \; f\text{'}(x)\; =\; 0$ at either of these points.

Now, by assumption, $f$ is continuous on and so $\backslash \; f\text{'}(x)\; =\; 0$ at every point of $\backslash \; (a,b)$.

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

For $\backslash \; x\_1$ less than $\backslash \; x$, $\backslash \; f(x)-f(x\_1)\; \backslash over\; \{x-x\_1\}$ is non-negative, since $x$ is a maximum. Thus the limit $\backslash \; \backslash lim\_\{x\_1\; \backslash rightarrow\; x^-\}$ is non-negative. (Note that we assume that $\backslash \; f$ is differentiable to guarantee that the left-hand and right-hand derivatives exist; it does not follow from the other assumptions).

For $\backslash \; x\_2$ greater than $\backslash \; x$, $\backslash \; f(x)-f(x\_2)\; \backslash over\; \{x-x\_2\}$ is non-positive. Thus $\backslash \; \backslash lim\_\{x\_2\; \backslash rightarrow\; x^+\}$ is non-positive.

Finally, since $\backslash \; f$ is differentiable at $\backslash \; x$, these two limits must be equal and hence are both 0. This implies that $\backslash \; f\text{'}(x)\; =\; 0$.

Generalization

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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 : 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 ±∞.

See also

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• Mean value theorem
• Extreme value theorem
• Intermediate value theorem
• Linear interpolation

External links

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• Rolle's and Mean Value Theorems at cut-the-knot

Calculus | Mathematical theorems

Teorema de Rolle | Rolles sætning | Satz von Rolle | Teorema de Rolle | Théorème de Rolle | 롤의 정리 | Teorema di Rolle | משפט רול | Rolle tétele | Stelling van Rolle | ロルの定理 | Twierdzenie Rolle'a | Теорема Ролля | Rollen lause | Rolles sats | 罗尔定理