The Mean Value Theorem

Formal statement

The function $f$ attains the slope of the secant between $a$ and $b$ as the derivative at the point $\xi \in (a,b)$.

It is also possible that there are multiple tangents parallel to the secant.

Let $f:[a,b]\to \mathbb {R}$ 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 $f:[a,b]\to \mathbb {R}$ 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 $x\mapsto x^{1/3}$, 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 Template:Nowrap Then

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

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

These formal statements are also known as Lagrange's Mean Value Theorem.

Proof

The expression ${\textstyle {\frac {f(b)-f(a)}{b-a}}}$ 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 on the curve lying between the end-points of the chord such that the tangent of the curve 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

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

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,

${\begin{aligned}&g'(x)=f'(x)-r\\&g'(c)=0\\&g'(c)=f'(c)-r=0\\&\Rightarrow f'(c)=r={\frac {f(b)-f(a)}{b-a}}\end{aligned}}$

Implications

Theorem 1: 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 Template:Nowrap. 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.
• The differentiability of f can be relaxed to one-sided differentiability, a proof given in the article on semi-differentiability.

Theorem 2': If f'(x) = g(x) for all x in an interval (a, b) of the domain of these functions, then f - g is constant or f = g + c where c is a constant on (a, b). === Proof: Let F = f − g, then F' = f' − g' = 0 on the interval (a, b), so the above theorem 1 tells that F = f − g is a constant c or f = g + c.

Theorem 3: If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is F(x) + c where c is an constant. === Proof: It is directly derived from the above theorem 2.

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 the 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 $c\in (a,b)$, such that

Geometrical meaning of Cauchy's theorem

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

Of course, if $g(a)\neq g(b)$ and $g'(c)\neq 0$, 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

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

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

$t\mapsto \left(t^{3},1-t^{2}\right),$

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 $g(a)\neq g(b)$. Define $h(x)=f(x)-rg(x)$, where $r$ is fixed in such a way that $h(a)=h(b)$, namely

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

  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.

Resources

• Mean Value Theorem, Wikipedia