In calculus, the mean value theorem states, roughly, that given an arc of a smooth continuous (derivatives) curve, there is at least one point on that arc at which the derivative (slope) of the curve is equal (parallel) to the "average" derivative of the arc. It is used to prove theorems that make global conclusions about a function on an interval starting from local hypotheses about derivatives at points of the interval. I like to share this Anti derivative with you all through my article.
Formal Statement of Mean Value Theorem Derivatives:
Let f: [x, y] → R be a continuous function on the closed interval [x, y] , and differentiable on the open interval (x, y), where x < y. Then there exists some z in (x, y) such that,
f'(z) = `(f(y) - f(x))/(y - x)`
The mean value theorem is a generalization of Rolle's Theorem, which assumes f(x) = f (y), so that the right-hand side above is zero.
Only one needs to assume that f: [x, y] → R is continuous on [x, y] , and that for every m in (x, y) the limit,
`lim_(h ->0) (f(m + h) - f(m))/(h)`
exists as a finite number or equals + ∞ or − ∞. If finite, that limit equals f′(m). An example where this version of the theorem applies is given by the real-valued cube root function mapping m to m1/3, whose derivative tends to infinity at the origin.
Note that the theorem is false if a differentiable function is complex-valued instead of real-valued.
For example, define f(m) = eim for all real m. Then
f (2π) − f(0) = 0 = 0(2π − 0)
while, |f′(m)| = 1.
- Source Wikipedia
Proof of Mean Value Theorem Derivatives:
Let g(m) = f(m) − rm, where r is a constant, because f is continuous on the closed interval [x, y] and differentiable on the open interval (x, y), now we want to select r, so as g satisfies the conditions of the Theorem,
g (x) = g (y) `hArr` f (x) − rx = f (y) − ry
`hArr` ry − rx = f (y) − f(x)
`hArr` r(y − x) = f (y) − f(x)
`hArr` r = `(f(y) - f(x))/(y - x)`
By Rolle's theorem, g is continuous on the closed interval [a, b] and g(a) = g(b), there is some c in (a, b) for which g′(c) = 0, and it follows from the equality g(x) = f(x) − rx that,
f' (c) = g' (c) + r = 0 + r = `(f(b) - f(a))/(b - a)`
Formal Statement of Mean Value Theorem Derivatives:
Let f: [x, y] → R be a continuous function on the closed interval [x, y] , and differentiable on the open interval (x, y), where x < y. Then there exists some z in (x, y) such that,
f'(z) = `(f(y) - f(x))/(y - x)`
The mean value theorem is a generalization of Rolle's Theorem, which assumes f(x) = f (y), so that the right-hand side above is zero.
Only one needs to assume that f: [x, y] → R is continuous on [x, y] , and that for every m in (x, y) the limit,
`lim_(h ->0) (f(m + h) - f(m))/(h)`
exists as a finite number or equals + ∞ or − ∞. If finite, that limit equals f′(m). An example where this version of the theorem applies is given by the real-valued cube root function mapping m to m1/3, whose derivative tends to infinity at the origin.
Note that the theorem is false if a differentiable function is complex-valued instead of real-valued.
For example, define f(m) = eim for all real m. Then
f (2π) − f(0) = 0 = 0(2π − 0)
while, |f′(m)| = 1.
- Source Wikipedia
Proof of Mean Value Theorem Derivatives:
Let g(m) = f(m) − rm, where r is a constant, because f is continuous on the closed interval [x, y] and differentiable on the open interval (x, y), now we want to select r, so as g satisfies the conditions of the Theorem,
g (x) = g (y) `hArr` f (x) − rx = f (y) − ry
`hArr` ry − rx = f (y) − f(x)
`hArr` r(y − x) = f (y) − f(x)
`hArr` r = `(f(y) - f(x))/(y - x)`
By Rolle's theorem, g is continuous on the closed interval [a, b] and g(a) = g(b), there is some c in (a, b) for which g′(c) = 0, and it follows from the equality g(x) = f(x) − rx that,
f' (c) = g' (c) + r = 0 + r = `(f(b) - f(a))/(b - a)`
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