How does Rolle's theorem differ from the mean value theorem?

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neela | High School Teacher | (Level 3) Valedictorian

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Rolle's theorem states that for a continuous real valued function f(x) in a closed interval  (a, b) and differentiable in the open interval (a,b), b> a with f(a) = f(b) , there exists a point c where f'(c) = 0.

The mean value theorem states that for a continuous real valued function f(x) in the closed interval (a,b), and differentiablen the open interval (a,b), b > a there exists a point c in (a,b)  for which

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

If we examine both theorems, it looks Rolle's theorem is a special case of mean value theorem. 

In Rolle's f(a) = f(b) . In mean value theorem f(a) and f(b) are not equal. 

In rolle's theorem f'(c) = 0 . The curve has a tangent || to x axis.

In the mean value theorem the slope f'(c) is not zero. f'(c) is the slope of the tangent at the point c and this tangent is not parallel to x axis. But this tangent is || to the line joining (a , f(a) ) and the point (b , f(b)) the end points of the curve.

Hope this helps.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Rolle's theorem represents the particular case of Mean Value Theorem.

The Mean Value Theorem is valid for continuous functions on closed intervals [a;b] and differentiable on the open interval (a;b).

The Mean Value Theorem states that there is a value c that belongs to the interval [a;b], so that:

f'(c) = [f(b) - f(a)]/(b-a) (1)

The Rolle's theorem states that f(a) = f(b) (2).

We'll substitute (2) in (1):

f'(c) = [f(a) - f(a)]/(b-a)

f'(c) = 0/(b-a)

f'(c) = 0

f'(c) gives the slope of the tangent to the curve of f(x), at the point (c,f(c)).

Since the slope of the tangent line is 0, that means that the tangent line is parallel to x axis and the point (c,f(c)) is a local extreme point.

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