Mean value theorem

For any function that is continuous on $[a,b]$ and differentiable on $(a,b)$ there exists some $c$ in the interval $(a,b)$ such that the secant joining the endpoints of the interval $[a,b]$ is parallel to the tangent at $c$ .

In mathematics, the mean value theorem states, roughly: that given a planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints.

The theorem is used to prove global statements about a function on an interval starting from local hypotheses about derivatives at points of the interval.

More precisely, if a function $f$ is continuous on the closed interval $[a,b]$ , where $a<b$ , and differentiable on the open interval $(a,b)$ , then there exists a point $c$ in $(a,b)$ such that:^[1]

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

A special case of this theorem was first described by Parameshvara (1370–1460), from the Kerala school of astronomy and mathematics in India, in his commentaries on Govindasvāmi and Bhaskara II.^[2] The mean value theorem in its modern form was later stated by Augustin Louis Cauchy (1789–1857). It is one of the most important results in differential calculus, as well as one of the most important theorems in mathematical analysis, and is useful in proving the fundamental theorem of calculus. The mean value theorem follows from the more specific statement of Rolle's theorem, and can be used to prove the more general statement of Taylor's theorem (with Lagrange form of the remainder term).

Formal statement

Let $f:[a,b]\to\R$ be a continuous function on the closed interval $[a,b]$ , and differentiable on the open interval $(a,b)$, where $a<b$ . Then there exists some $c$ in $(a,b)$ such that

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

The mean value theorem is a generalization of Rolle's theorem, which assumes $f(a)=f(b)$, so that the right-hand side above is zero.

The mean value theorem is still valid in a slightly more general setting. One only needs to assume that $f:[a,b]\to\R$ is continuous on $[a,b]$ , and that for every $x$ in $(a,b)$ the limit

$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

exists as a finite number or equals $\infty$ or $-\infty$ . If finite, that limit equals $f'(x)$ . An example where this version of the theorem applies is given by the real-valued cube root function mapping $x \to x^\frac13$ , whose derivative tends to infinity at the origin.

Note that the theorem, as stated, is false if a differentiable function is complex-valued instead of real-valued. For example, define $f(x)=e^{xi}$ for all real $x$ . Then

$f(2\pi)-f(0)=0=0(2\pi-0)$

while $f'(x)\ne 0$ for any real $x$ .

Proof

The expression $\frac{f(b)-f(a)}{b-a}$ gives the slope of the line joining the points $(a,f(a))$ and $(b,f(b))$ , which is a chord of the graph of $f$ , while $f'(x)$ gives the slope of the tangent to the curve at the point $(x,f(x))$ . Thus the Mean value theorem says that given any chord of a smooth curve, we can find a point lying between the end-points of the chord such that the tangent at that point is parallel to the chord. The following proof illustrates this idea.

Define $g(x)=f(x)-rx$ , where $r$ is a constant. Since $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , the same is true for $g$ . We now want to choose $r$ so that $g$ satisfies the conditions of Rolle's theorem. Namely

$\begin{align}g(a)=g(b)&\iff f(a)-ra=f(b)-rb\\ &\iff r(b-a)=f(b)-f(a) \\&\iff r=\frac{f(b)-f(a)}{b-a}\cdot\end{align}$

By Rolle's theorem, since $g$ is differentiable 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,

$f'(c)=g'(c)+r=0+r=\frac{f(b)-f(a)}{b-a}$

as required.

A simple application

Assume that f is a continuous, real-valued function, defined on an arbitrary interval I of the real line. If the derivative of f at every interior point of the interval I exists and is zero, then f is constant in the interior.

Proof: Assume the derivative of f at every interior point of the interval I exists and is zero. Let (a, b) be an arbitrary open interval in I. By the mean value theorem, there exists a point c in (a,b) such that

$0=f'(c)=\frac{f(b)-f(a)}{b-a}.$

This implies that f(a) = f(b). Thus, f is constant on the interior of I and thus is constant on I by continuity. (See below for a multivariable version of this result.)

Remarks:

• Only continuity of f, not differentiability, is needed at the endpoints of the interval I. No hypothesis of continuity needs to be stated if I is an open interval, since the existence of a derivative at a point implies the continuity at this point. (See the section continuity and differentiability of the article derivative.)
• The differentiability of f can be relaxed to one-sided differentiability, a proof given in the article on semi-differentiability.

Cauchy's mean value theorem

Cauchy's mean value theorem, also known as the extended mean value theorem, is a generalization of the mean value theorem. It states: If functions f and g are both continuous on the closed interval [a, b], and differentiable on the open interval (a, b), then there exists some c ∈ (a, b), such that

Geometrical meaning of Cauchy's theorem

$(f(b)-f(a))g'(c)=(g(b)-g(a))f'(c).$

Of course, if g(a) ≠ g(b) and if g′(c) ≠ 0, this is equivalent to:

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

Geometrically, this means that there is some tangent to the graph of the curve

$\begin{cases}[a,b] \to \mathbf{R}^2\\t\mapsto (f(t),g(t))\end{cases}$

which is parallel to the line defined by the points (f(a), g(a)) and (f(b), g(b)). However Cauchy's theorem does not claim the existence of such a tangent in all cases where (f(a), g(a)) and (f(b), g(b)) are distinct points, since it might be satisfied only for some value c with f′(c) = g′(c) = 0, in other words a value for which the mentioned curve is stationary; in such points no tangent to the curve is likely to be defined at all. An example of this situation is the curve given by

$t\mapsto(t^3,1-t^2),$

which on the interval [−1, 1] goes from the point (−1, 0) to (1, 0), yet never has a horizontal tangent; however it has a stationary point (in fact a cusp) at t = 0.

Cauchy's mean value theorem can be used to prove l'Hôpital's rule. The mean value theorem is the special case of Cauchy's mean value theorem when g(t) = t.

Proof of Cauchy's mean value theorem

The proof of Cauchy's mean value theorem is based on the same idea as the proof of the mean value theorem.

• Suppose g(a) ≠ g(b). Define h(x) = f(x) − rg(x), where r is fixed in such a way that h(a) = h(b), namely

$\begin{align}h(a)=h(b)&\iff f(a)-rg(a)=f(b)-rg(b)\\ &\iff r (g(b)-g(a))=f(b)-f(a)\\ &\iff r=\frac{f(b)-f(a)}{g(b)-g(a)}.\end{align}$

Since f and g are continuous on [a, b] and differentiable on (a, b), the same is true for h. All in all, h satisfies the conditions of Rolle's theorem: consequently, there is some c in (a, b) for which h′(c) = 0. Now using the definition of h we have:

$0=h'(c)=f'(c)-rg'(c) = f'(c)- \left (\frac{f(b)-f(a)}{g(b)-g(a)} \right ) g'(c).$

Therefore:

$f'(c)= \frac{f(b)-f(a)}{g(b)-g(a)} g'(c),$

which implies the result.

• If g(a) = g(b), then, applying Rolle's theorem to g, it follows that there exists c in (a, b) for which g′(c) = 0. Using this choice of c, Cauchy's mean value theorem (trivially) holds.

Generalization for determinants

Assume that $f, g$ and $h$ are differentiable functions on $(a,b)$ that are continuous on $[a,b]$. Define

$D(x)=\left |\begin{array}{ccc}f(x) & g(x)& h(x)\\ f(a) & g(a) & h(a)\\ f(b) & g(b)& h(b)\end{array}\right|$

There exists $c\in(a,b)$ such that $D'(c)=0$.

Notice that

$D'(x)=\left |\begin{array}{ccc}f'(x) & g'(x)& h'(x)\\ f(a) & g(a) & h(a)\\ f(b) & g(b)& h(b)\end{array}\right|$

and if we place $h(x)=1$, we get Cauchy's mean value theorem. If we place $h(x)=1$ and $g(x)=x$ we get Lagrange's mean value theorem.

The proof of the generalization is quite simple: each of $D(a)$ and $D(b)$ are determinants with two identical rows, hence $D(a)=D(b)=0$. The Rolle's theorem implies that there exists $c\in (a,b)$ such that $D'(c)=0$.

Mean value theorem in several variables

The mean value theorem generalizes to real functions of multiple variables. The trick is to use parametrization to create a real function of one variable, and then apply the one-variable theorem.

Let $G$ be an open connected subset of $\R^n$ , and let $f:G\to\R$ be a differentiable function. Fix points $x,y\in G$ such that the interval $x\ y$ lies in $G$ , and define $g(t)=f\Big((1-t)x+ty\Big)$ . Since $g$ is a differentiable function in one variable, the mean value theorem gives:

$g(1)-g(0)=g'(c)$

for some $c$ between 0 and 1. But since $g(1)=f(y)$ and $g(0)=f(x)$ , computing $g'(c)$ explicitly we have:

$f(y)-f(x)=\nabla f\Big((1-c)x+cy\Big)\cdot (y-x)$

where $\nabla$ denotes a gradient and $\cdot$ a dot product. Note that this is an exact analog of the theorem in one variable (in the case $n=1$ this is the theorem in one variable). By the Cauchy-Schwarz inequality, the equation gives the estimate:

$\Bigg|f(y)-f(x)\Bigg|\le\Bigg|\nabla f\Big((1-c)x+cy\Big)\Bigg|\ \Big|y - x\Big|.$

In particular, when the partial derivatives of $f$ are bounded, $f$ is Lipschitz continuous (and therefore uniformly continuous). Note that $f$ is not assumed to be continuously differentiable nor continuous on the closure of $G$ . However, in the above, we used the chain rule so the existence of $\nabla f$ would not be sufficient.

As an application of the above, we prove that $f$ is constant if $G$ is open and connected and every partial derivative of $f$ is 0. Pick some point $x_0\in G$ , and let $g(x)=f(x)-f(x_0)$ . We want to show $g(x)=0$ for every $x\in G$ . For that, let $E=x\in G:g(x)=0$ . Then E is closed and nonempty. It is open too: for every $x\in E$ ,

$\Big|g(y)\Big|=\Bigg|g(y)-g(x)\Bigg|\le (0)\Big|y-x\Big|=0$

for every $y$ in some neighborhood of $x$ . (Here, it is crucial that $x$ and $y$ are sufficiently close to each other.) Since $G$ is connected, we conclude $E=G$ .

Remark that all arguments in the above are made in a coordinate-free manner; hence, they actually generalize to the case when $G$ is a subset of a Banach space.

Mean value theorem for vector-valued functions

There is no exact analog of the mean value theorem for vector-valued functions.

Jean Dieudonné in his classic treatise Foundations of Modern Analysis discards the mean value theorem and replaces it by mean inequality as the proof is not constructive and one cannot find the mean value and in applications one only needs mean inequality. Serge Lang in Analysis I uses the mean value theorem, in integral form, as an instant reflex but this use requires the continuity of the derivative. If one uses the Henstock-Kurzweil integral one can have the mean value theorem in integral form without the additional assumption that derivative should be continuous as every derivative is Henstock-Kurzweil integrable. The problem is roughly speaking the following: If f : U → R^m is a differentiable function (where U ⊂ Rⁿ is open) and if x + th, x, h ∈ Rⁿ, t ∈ [0, 1] is the line segment in question (lying inside U), then one can apply the above parametrization procedure to each of the component functions f_i (i = 1, ..., m) of f (in the above notation set y = x + h). In doing so one finds points x + t_ih on the line segment satisfying

$f_i(x+h) - f_i(x) = \nabla f_i (x + t_ih) \cdot h.$

But generally there will not be a single point x + t*h on the line segment satisfying

$f_i(x+h) - f_i(x) = \nabla f_i (x + t^* h) \cdot h.$

for all i simultaneously. For example define:

$\begin{cases} f : [0, 2 \pi] \to \mathbf{R}^2 \\ f(x) = (\cos(x), \sin(x)) \end{cases}$

Then f(2π) − f(0) = 0 ∈ R², but $f_1'(x)=-\sin (x)$ and $f_2'(x)=\cos (x)$ are never simultaneously zero as x ranges over [0, 2π].)

However a certain type of generalization of the mean value theorem to vector-valued functions is obtained as follows: Let f be a continuously differentiable real-valued function defined on an open interval I, and let x as well as x + h be points of I. The mean value theorem in one variable tells us that there exists some t* between 0 and 1 such that

$f(x+h)-f(x) = f'(x+t^*h)\cdot h.$

On the other hand, we have, by the fundamental theorem of calculus followed by a change of variables,

$f(x+h)-f(x) = \int_x^{x+h} f'(u)du = \left (\int_0^1 f'(x+th)\,dt\right)\cdot h.$

Thus, the value f′(x + t*h) at the particular point t* has been replaced by the mean value

$\int_0^1 f'(x+th)\,dt.$

This last version can be generalized to vector valued functions:

Lemma 1. Let U ⊂ Rⁿ be open, f : U → R^m continuously differentiable, and x ∈ U, h ∈ Rⁿ vectors such that the line segment x + th, 0 ≤ t ≤ 1 remains in U. Then we have:

$f(x+h)-f(x) = \left (\int_0^1 Df(x+th)\,dt\right)\cdot h,$

where Df denotes the Jacobian matrix of f and the integral of a matrix is to be understood componentwise.

Proof. Let f₁, ..., f_m denote the components of f and define:

$\begin{cases} g_i : [0,1] \to \mathbf{R} \\ g_i(t) = f_i (x +th) \end{cases}$

Then we have

$f_i(x+h)-f_i(x) = g_i(1)-g_i(0) =\int_0^1 g_i'(t)dt = \int_0^1 \left (\sum_{j=1}^n \frac{\partial f_i}{\partial x_j} (x+th)h_j\right)\,dt =\sum_{j=1}^n \left (\int_0^1 \frac{\partial f_i}{\partial x_j}(x+th)\,dt\right)h_j.$

The claim follows since Df is the matrix consisting of the components $\tfrac{\partial f_i}{\partial x_j}.$

Lemma 2. Let v : [a, b] → R^m be a continuous function defined on the interval [a, b] ⊂ R. Then we have

$\left \|\int_a^b v(t)\,dt\right\|\leqslant \int_a^b \|v(t)\|\,dt.$

Proof. Let u in R^m denote the value of the integral

$u:=\int_a^b v(t)\,dt.$

Now we have (using the Cauchy–Schwarz inequality):

$\|u\|^2 = \langle u,u \rangle =\left \langle \int_a^b v(t) dt,u \right\rangle = \int_a^b \langle v(t),u \rangle \,dt \leqslant \int_a^b \| v(t) \|\cdot \|u \|\,dt = \|u\| \int_a^b \|v(t)\|\,dt$

Now cancelling the norm of u from both ends gives us the desired inequality.

Mean Value Inequality. If the norm of Df(x + th) is bounded by some constant M for t in [0, 1], then

$\|f(x+h)-f(x)\| \leqslant M\|h\|.$

Proof. From Lemma 1 and 2 it follows that

$\|f(x+h)-f(x)\|=\left \|\int_0^1 (Df(x+th)\cdot h)\,dt\right\| \leqslant \int_0^1 \|Df(x+th)\| \cdot \|h\|\, dt \leqslant M\| h\|.$

Mean Value Theorems for Definite Integrals

First Mean Value Theorem for Definite Integrals

Let f : [a, b] → R be a continuous function. Then there exists c in (a, b) such that

$\int_a^b f(x) \, dx = f(c)(b - a).$

Since the mean value of f on [a, b] is defined as

$\frac{1}{b-a} \int_a^b f(x) \, dx,$

we can interpret the conclusion as f achieves its mean value at some c in (a, b).^[3]

In general, if f : [a, b] → R is continuous and g is an integrable function that does not change sign on [a, b], then there exists c in (a, b) such that

$\int_a^b f(x) g(x) \, dx = f(c) \int_a^b g(x) \, dx.$

Proof of the First Mean Value Theorem for Definite Integrals

Suppose f : [a, b] → R is continuous and g is a nonnegative integrable function on [a, b]. By the extreme value theorem, there exists m and M such that for each x in [a, b], $m\leqslant f(x) \leqslant M$ and $f[a,b] = [m, M]$. Since g is nonnegative,

$m \int_a^b g(x) \, dx \leqslant \int^b_a f(x)g(x) \, dx \leqslant M \int_a^b g(x) \, dx.$

Now let

$I = \int_a^b g(x) \, dx.$

If $I = 0$, we're done since

$0 \leqslant \int_a^b f(x) g(x)\, dx \leqslant 0$

means

$\int_a^b f(x)g(x)\, dx=0,$

so for any c in (a, b),

$\int_a^b f(x)g(x)\, dx = f(c) I = 0.$

If I ≠ 0, then

$m \leqslant \frac1I \int_a^b f(x)g(x)\,dx \leqslant M.$

By the intermediate value theorem, f attains every value of the interval [m, M], so for some c in [a, b]

$f(c) = \frac1I\int^b_a f(x) g(x) \, dx,$

that is,

$\int_a^b f(x) g(x) \, dx = f(c) \int_a^b g(x) \, dx.$

Finally, if g is negative on [a, b], then

$M \int_a^b g(x) \, dx \leqslant \int^b_a f(x)g(x) \, dx \leqslant m \int_a^b g(x) \, dx,$

and we still get the same result as above.

QED

Second Mean Value Theorem for Definite Integrals

There are various slightly different theorems called the second mean value theorem for definite integrals. A commonly found version is as follows:

If G : [a, b] → R is a positive monotonically decreasing function and φ : [a, b] → R is an integrable function, then there exists a number x in (a, b] such that

$\int_a^b G(t)\varphi(t)\,dt = G(a^+) \int_a^x \varphi(t)\,dt.$

Here $G(a^+)$ stands for ${\lim_{x\to a^+}G(x)}$, the existence of which follows from the conditions. Note that it is essential that the interval (a, b] contains b. A variant not having this requirement is:^[4]

If G : [a, b] → R is a monotonic (not necessarily decreasing and positive) function and φ : [a, b] → R is an integrable function, then there exists a number x in (a, b) such that

$\int_a^b G(t)\varphi(t)\,dt = G(a^+) \int_a^x \varphi(t)\,dt + G(b^-) \int_x^b \varphi(t)\,dt.$

Mean value theorem for integration fails for vector-valued functions

If the function $G$ returns a multi-dimensional vector, then the MVT for integration is not true, even if the domain of $G$ is also multi-dimensional.

For example, consider the following 2-dimensional function defined on an $n$-dimensional cube:

$\begin{cases} G: [0,2\pi]^n \to \mathbb{R}^2 \\ G(x_1,\cdots,x_n)=\left (\sin(x_1+\cdots+x_n), \cos(x_1+\cdots+x_n)\right)\end{cases}$

Then, by symmetry it is easy to see that the mean value of $G$ over its domain is (0,0):

$\int_{[0,2\pi]^n} G(x_1,\cdots,x_n) dx_1 \cdots dx_n = (0,0)$

However, there is no point in which $G=(0,0)$, because $|G|=1$ everywhere.

A probabilistic analogue of the mean value theorem

Let X and Y be non-negative random variables such that E[X] < E[Y] < ∞ and $X\leqslant_{st} Y$ (i.e. X is smaller than Y in the usual stochastic order). Then there exists an absolutely continuous non-negative random variable Z having probability density function

$f_Z(x)={\Pr(Y>x)-\Pr(X>x)\over {\rm E}[Y]-{\rm E}[X]}\,, \qquad x\geqslant 0.$

Let g be a measurable and differentiable function such that E[g(X)], E[g(Y)] < ∞, and let its derivative g′ be measurable and Riemann-integrable on the interval [x, y] for all y ≥ x ≥ 0. Then, E[g′(Z)] is finite and^[5]

${\rm E}[g(Y)]-{\rm E}[g(X)]={\rm E}[g'(Z)]\,[{\rm E}(Y)-{\rm E}(X)].$

Generalization in complex analysis

As noted above, the theorem does not hold for differentiable complex-valued functions. Instead, a generalization of the theorem is stated such:^[6]

Let f : Ω → C be a holomorphic function on the open convex set Ω, and let a and b be distinct points in Ω. Then there exist points u, v on L_ab (the line segment from a to b) such that

$\mathrm{Re}(f'(u)) = \mathrm{Re}\left ( \frac{f(b)-f(a)}{b-a} \right),$

$\mathrm{Im}(f'(v)) = \mathrm{Im}\left ( \frac{f(b)-f(a)}{b-a} \right).$

Where Re() is the Real part and Im() is the Imaginary part of a complex-valued function.

See also

• Newmark-beta method
• Mean value theorem (divided differences)

Notes

1. Weisstein, Eric. "Mean-Value Theorem". MathWorld. Wolfram Research. Retrieved 24 March 2011.
2. J. J. O'Connor and E. F. Robertson (2000). Paramesvara, MacTutor History of Mathematics archive.
3. Michael Comenetz (2002). Calculus: The Elements. World Scientific. p. 159. ISBN 978-981-02-4904-5.
4. Hobson, E. W. (1909). "On the Second Mean-Value Theorem of the Integral Calculus". Proc. London Math Soc. S2–7 (1): 14–23. doi:10.1112/plms/s2-7.1.14. MR 1575669.
5. Di Crescenzo, A. (1999). "A Probabilistic Analogue of the Mean Value Theorem and Its Applications to Reliability Theory". J. Appl. Prob. 36 (3): 706–719. JSTOR 3215435.
6. "Complex Mean-Value Theorem". PlanetMath. PlanetMath.

External links

• Hazewinkel, Michiel, ed. (2001), "Cauchy theorem", Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4
• PlanetMath: Mean-Value Theorem
• Weisstein, Eric W., "Mean value theorem", MathWorld.
• Weisstein, Eric W., "Cauchy's Mean-Value Theorem", MathWorld.
• "Mean Value Theorem: Intuition behind the Mean Value Theorem" at the Khan Academy