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Mathematics / mean-value-theorem

Mean Value Theorem

The mean value theorem states that a differentiable real function attains its average rate of change at some interior point of an interval.

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The mean value theorem is a fundamental result of calculus connecting the average change of a function over an interval with its instantaneous rate of change. For a real-valued function continuous on a closed interval and differentiable in its interior, the theorem guarantees an interior point where the derivative equals the slope between the endpoints. It underlies many results about the behavior of differentiable functions. (openstax.org)

Statement and interpretation

Let a<ba<b, and let f:[a,b]→Rf:[a,b]\to\mathbb R satisfy:

  1. ff is continuous on [a,b][a,b].
  2. ff is differentiable on (a,b)(a,b).

Then there exists at least one c∈(a,b)c\in(a,b) such that

f′(c)=f(b)−f(a)b−a.f'(c)=\frac{f(b)-f(a)}{b-a}.

Equivalently,

f(b)−f(a)=f′(c)(b−a).f(b)-f(a)=f'(c)(b-a).

Geometrically, the tangent to the graph at cc is parallel to the secant joining (a,f(a))(a,f(a)) and (b,f(b))(b,f(b)). If f(t)f(t) represents position along a straight line, some instantaneous velocity equals the average velocity over the interval. (math.mit.edu)

The theorem asserts existence, not uniqueness, and does not say that cc is the midpoint. It requires neither derivatives at the endpoints nor continuity of f′f'. (openstax.org)

For example, take f(x)=x2f(x)=x^2. Direct calculation gives

f(b)−f(a)b−a=a+b.\frac{f(b)-f(a)}{b-a}=a+b.

Since f′(x)=2xf'(x)=2x, the required point is c=(a+b)/2c=(a+b)/2. This midpoint result is specific to this example: for f(x)=x3f(x)=x^3 on [0,1][0,1], the equation 3c2=13c^2=1 instead gives c=1/3c=1/\sqrt3.

Rolle’s theorem and proof

The standard proof uses Rolle’s theorem: under the same continuity and differentiability assumptions, if f(a)=f(b)f(a)=f(b), then f′(c)=0f'(c)=0 at some interior point.

Rolle’s theorem follows from the extreme value theorem. A nonconstant continuous function with equal endpoint values must attain a maximum or minimum in the interior; differentiability forces its derivative there to vanish. A constant function has derivative zero everywhere. (metaphor.ethz.ch)

To obtain the mean value theorem, subtract the secant line from ff. Define

h(x)=f(x)−f(a)−f(b)−f(a)b−a(x−a).h(x)=f(x)-f(a) -\frac{f(b)-f(a)}{b-a}(x-a).

Then h(a)=h(b)=0h(a)=h(b)=0, and hh satisfies Rolle’s hypotheses. Consequently, some c∈(a,b)c\in(a,b) satisfies

0=h′(c)=f′(c)−f(b)−f(a)b−a,0=h'(c) =f'(c)-\frac{f(b)-f(a)}{b-a},

which establishes the claimed equality. (metaphor.ethz.ch)

Consequences and applications

Applying the theorem on each subinterval gives several basic consequences:

  • Zero derivative: if f′=0f'=0 throughout an interval, ff is constant there.
  • Equal derivatives: two functions with identical derivatives differ by a constant. This explains the constant of integration in an antiderivative.
  • Monotonicity: a positive derivative implies strict increase; a negative derivative implies strict decrease. Nonnegative and nonpositive derivatives imply the corresponding non-strict forms of monotonicity. (openstax.org)

Another direct consequence is a bound on changes. If ∣f′(t)∣≤M|f'(t)|\le M between xx and yy, the theorem gives

∣f(y)−f(x)∣≤M∣y−x∣.|f(y)-f(x)|\le M|y-x|.

Thus a uniform derivative bound implies Lipschitz continuity. Such estimates also have vector-valued analogues, even where the exact mean value equality fails. (maths.tcd.ie)

For example, because ∣cos⁡t∣≤1|\cos t|\le1, applying the theorem to sine yields

∣sin⁡x−sin⁡y∣≤∣x−y∣.|\sin x-\sin y|\le|x-y|.

The theorem also supplies an exact first-order increment formula,

f(x+h)=f(x)+hf′(x+θh),0<θ<1,f(x+h)=f(x)+hf'(x+\theta h), \qquad 0<\theta<1,

provided the hypotheses hold between xx and x+hx+h. Higher-order analogues occur in Taylor’s theorem, which controls the error of polynomial approximation. (ocw.mit.edu)

Why the hypotheses matter

Continuity at the endpoints cannot simply be omitted. Consider the directly constructed example

f(x)={0,0≤x<1,1,x=1.f(x)= \begin{cases} 0,&0\le x<1,\\ 1,&x=1. \end{cases}

Its derivative is zero everywhere in (0,1)(0,1), but its endpoint slope is 11; no required point exists.

Interior differentiability is also essential. For f(x)=∣x∣f(x)=|x| on [−1,1][-1,1], the endpoint slope is zero. Wherever the derivative exists, however, it is either −1-1 or 11, and at zero it is undefined. The mean value conclusion therefore fails. These examples illustrate why both assumptions appear in the theorem. (openstax.org)

The assumptions are sufficient, not necessary for an individual function to happen to satisfy the equality. Nor does the theorem permit one fixed cc to represent every subinterval.

Cauchy’s generalized mean value theorem

Cauchy’s mean value theorem concerns two functions ff and gg, both continuous on [a,b][a,b] and differentiable on (a,b)(a,b). It guarantees some c∈(a,b)c\in(a,b) such that

[f(b)−f(a)]g′(c)=[g(b)−g(a)]f′(c).[f(b)-f(a)]g'(c) =[g(b)-g(a)]f'(c).

If g′g' is nowhere zero in the interior, Rolle’s theorem ensures g(b)≠g(a)g(b)\ne g(a), and division gives

f(b)−f(a)g(b)−g(a)=f′(c)g′(c).\frac{f(b)-f(a)}{g(b)-g(a)} =\frac{f'(c)}{g'(c)}.

Taking g(x)=xg(x)=x recovers the ordinary theorem. The generalized result is used in proofs of L’Hôpital’s rule for evaluating certain quotient limits. The cross-multiplied statement is the more general form because it does not require division by possibly zero quantities. (ocw.mit.edu)

Integral and multivariable versions

The mean value theorem for integrals states that a continuous real-valued function on [a,b][a,b] attains its integral average: some c∈[a,b]c\in[a,b] satisfies

∫abf(x) dx=f(c)(b−a).\int_a^b f(x)\,dx=f(c)(b-a).

Unlike the differential theorem, this statement concerns function values rather than derivatives and requires no differentiability. When f′f' is continuous, the fundamental theorem of calculus relates the two through

f(b)−f(a)=∫abf′(x) dx.f(b)-f(a)=\int_a^b f'(x)\,dx.

This integral argument uses stronger assumptions than the ordinary differential theorem. (openstax.org)

For a differentiable scalar function of several real variables, applying the one-dimensional theorem along a line segment gives

f(y)−f(x)=∇f(c)⋅(y−x),f(y)-f(x)=\nabla f(c)\cdot(y-x),

where c=x+θ(y−x)c=x+\theta(y-x), 0<θ<10<\theta<1, provided the segment lies in the function’s open domain. Here ∇f\nabla f is the gradient. (maths.tcd.ie)

An exact equality with a single intermediate derivative generally fails for vector-valued functions. For

F(t)=(cos⁡t,sin⁡t),0≤t≤2π,F(t)=(\cos t,\sin t),\qquad 0\le t\le2\pi,

the endpoints coincide, but F′(t)=(−sin⁡t,cos⁡t)F'(t)=(-\sin t,\cos t) never vanishes. Norm inequalities replace the scalar equality; for a continuously differentiable function taking values in a Banach space,

∥F(b)−F(a)∥≤(b−a)sup⁡a≤t≤b∥F′(t)∥.\|F(b)-F(a)\| \le(b-a)\sup_{a\le t\le b}\|F'(t)\|.

(maths.tcd.ie)

Historical development

Michel Rolle established a polynomial precursor of Rolle’s theorem in 1691. The differential mean value theorem developed through work associated with Lagrange and Augustin-Louis Cauchy; Cauchy published a precise formulation and proof in his 1823 lectures on infinitesimal calculus. Its modern formulation belongs to the nineteenth-century development of rigorous analysis. (math.camden.rutgers.edu)

References

  1. 4 The Mean Value Theorem - Calculus Volume 1openstax.org
  2. 1 Statement and Geometric Interpretationmath.mit.edu
  3. THE MEAN VALUE THEOREMmetaphor.ethz.ch
  4. 100B S25 Lecture 16: Rolle’s Theorem; Mean Theorem; L’Hôpital’s Rule; Taylor Expansionocw.mit.edu
  5. 3 The Fundamental Theorem of Calculus - Calculus Volume 1openstax.org
  6. The Mean Value Theorem for Vector Valued Functions: A Simple Proofmaths.tcd.ie
  7. Vector-valued integralsmath.ucdavis.edu
  8. Real Analysis: Mean Value Theorems for Derivativesmath.camden.rutgers.edu