Advanced Analysis

The Mean Value Theorem

The Mean Value Theorem is perhaps the most important and useful property of differentiable functions. Informally, it tells us that for differentiable functions, every secant line to the graph is parallel to an actual tangent line. In practice, the MVT is a bridge which allows us to deduce global properties (e.g., monotonicity) of differentiable functions from local properties (e.g., that \(f' > 0\) at each point).

1. Preparation: Local Inequalities Near Points of Differentiability and Rolle's Theorem

Lemma
Suppose \(f : (a,b) \rightarrow {\mathbb R}\) is differentiable at a point \(c \in (a,b)\).
  • If \(f'(c) > 0\), then there is an interval \(I\) centered at \(c\) such that \(f(x) > f(c)\) for all \(x \in I\) with \(x > c\) and \(f(x) < f(c)\) for all \(x \in I\) with \(x < c\).
  • If \(f'(c) < 0\), then there is an interval \(I\) centered at \(c\) such that \(f(x) < f(c)\) for all \(x \in I\) with \(x > c\) and \(f(x) > f(c)\) for all \(x \in I\) with \(x < c\).
Proof
Since \(|f'(c)| > 0\), there exists \(\delta > 0\) such that \(0 < |x-c| < \delta\) and \(x \in (a,b)\) implies
\[ \left| \frac{f(x) - f(c)}{x-c} - f'(c) \right| < \frac{|f'(c)|}{2}. \]
Without loss of generality, we may assume that \(\delta < \min\{ |c-a|, |c-b|\}\) so that \((c-\delta,c+\delta) \subset (a,b)\). Assuming \(f'(c) > 0\), on \(0 < |x-c| <\delta\) we have
\[{}\frac{f(x) - f(c)}{x-c} - f'(c) > - \frac{f'(c)}{2}{}\]
\[{}\Rightarrow \frac{f(x) - f(c)}{x-c} > \frac{f'(c)}{2}.{}\]
In particular, the difference quotient will be strictly positive, meaning that \(f(x) > f(c)\) when \(x > c\) and \(f(x) < f(c)\) when \(x < c\). If instead \(f'(c) < 0\), we have
\[{}\frac{f(x) - f(c)}{x-c} - f'(c) < -\frac{f'(c)}{2}{}\]
\[{}\Rightarrow \frac{f(x) - f(c)}{x-c} < \frac{f'(c)}{2}.{}\]
In this case we get exactly the opposite conclusion, namely, that the difference quotients are strictly negative and consequently the order relation between \(f(x)\) and \(f(c)\) is opposite the relation of \(x\) and \(c\).
Theorem 1 (Rolle's Theorem)
Let \(a\) and \(b\) be real numbers with \(a < b\). Suppose \(f : [a,b] \rightarrow {\mathbb R}\) is continuous on \([a,b]\) and differentiable on \((a,b)\). If \(f(a) = f(b)\), then there exists \(c \in (a,b)\) such that \(f'(c) = 0\).
Proof
By the Extreme Value Theorem, there exist points \(x_i\) and \(x_s\) in \([a,b]\) such that
\[f(x_i) = \inf_{x \in [a,b]} f(x)\]
and
\[f(x_s) = \sup_{x \in [a,b]} f(x).\]
If \(f(x_i) = f(x_s)\), then \(f\) must be constant on \([a,b]\), since for any \(x \in [a,b]\) we would have \(f(x_i) \leq f(x) \leq f(x_s) = f(x_i)\). Knowing that \(f\) is constant then implies the much stronger statement that \(f'(c) = 0\) for all \(c \in (a,b)\).

Having reached the desired conclusion when \(f(x_s) = f(x_i)\), we may next assume without loss of generality that \(f(x_i) < f(x_s)\). In particular, at least one of \(f(x_i)\) and \(f(x_s)\) differs from the boundary value \(f(a)\). If, for example, \(f(x_i) \neq f(a)\), then \(x_i\) certainly cannot equal \(a\), nor can it equal \(b\), since \(f(b) = f(a)\). Likewise if \(f(x_s) \neq f(a)\), then \(x_s\) is not equal to \(a\) or to \(b\). So we can conclude that at least one of the points \(x_i\) and \(x_s\) belongs to \((a,b)\). Let \(c := x_i\) if \(x_i \in (a,b)\) and let \(c := x_s\) otherwise. By the lemma above, it must be the case that \(f'(c) = 0\); for if not, then there would be points \(c_1\) and \(c_2\) in some neighborhood of \(c\) for which \(f(c_1) < f(c) < f(c_2)\), which would mean that \(f(c)\) could be neither an upper bound nor a lower bound for \(f(x)\) over all \(x \in [a,b]\) (and by our choice of \(c\) we know that \(f(c)\) must, in fact, be one or the other).
In the proof of Rolle's Theorem, we made a fundamental observation: If \(f\) is continuous on \([a,b]\), differentiable on \((a,b)\) and if either of its maximum value or minimum values is not attained at the endpoints, then there is a point \(c \in (a,b)\) at which \(f'(c) = 0\). We can use this observation to prove an interesting property of derivatives themselves:
Theorem 2 (Intermediate Value Property of Derivatives)
Suppose that \(f\) is a differentiable function on some interval \(I\) and that \(a,b\) in \(I\) have the property that \(f'(a) \neq f'(b)\). Then for any \(m\) strictly between \(f'(a)\) and \(f'(b)\), there is some \(c\) strictly between \(a\) and \(b\) such that \(f'(c) = m\).
Proof
By symmetry, it may be assumed that \(a < b\). Consider the function \(g(x) = f(x) - m(x)\). By assumption one of \(g'(a)\) and \(g'(b)\) is positive and the other is negative. Suppose first that \(g'(a) > 0 > g'(b)\). By the lemma above, the maximum of \(g\) on \([a,b]\) cannot be attained at \(a\) because \(g(x) > g(a)\) for all \(x > a\) sufficiently close to \(a\). Likewise the maximum of \(g\) on \([a,b]\) cannot occur at \(b\) because \(g(x) > g(b)\) for all \(x < b\) sufficiently close to \(b\). Thus the maximum of \(g\) on \([a,b]\) is attained somewhere in \((a,b)\) and has derivative \(0\) there. Thus \(0 = g'(c) = f'(c) - m\). In the remaining case \(g'(a) < 0 < g'(b)\), the minimum of \(g\) on \([a,b]\) cannot be attained at either endpoint, so once again, there is a point \(c \in (a,b)\) at which \(g'(c) = 0\).

2. The Mean Value Theorem and Generalized Mean Value Theorem

The picture below illustrates the Mean Value Theorem. We see a blue secant line joining points \((a,f(a))\) to \((b,f(b))\) on the graph of some differentiable function \(y = f(x)\) along with a parallel purple line which is tangent to the graph at \((c,f(c))\) for some point \(c\) strictly between \(a\) and \(b\) on the horizontal axis.
Figure. Illustration of the Mean Value Theorem
Corollary (Mean Value Theorem)
Suppose \(f : [a,b] \rightarrow {\mathbb R}\) is continuous on \([a,b]\) and differentiable on \((a,b)\). Then there exists \(c \in (a,b)\) such that
\[ f(b) - f(a) = f'(c) (b-a). \]
Proof
Let
\[ g(x) := f(x) - (x-a) \frac{f(b) - f(a)}{b-a}. \]
Since \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), the function \(g\) will have these same properties (since \(g - f\) is trivially continuous and differentiable everywhere). Moreover, \(g(a) = f(a)\) and
\[{} g(b){}\]
\[{}= f(b) - (b-a)\frac{f(b)-f(a)}{b-a}{}\]
\[{}= f(a) = g(a),{}\]
so by Rolle's Theorem, there is a point \(c \in (a,b)\) for which \(g'(c) = 0\), meaning that
\[ 0 = f'(c) - \frac{f(b) - f(a)}{b-a}, \]
exactly as desired.
The single most important implication of the Mean Value Theorem is that if \(f\) is a differentiable function on some open interval and \(f' > 0\) at every point of the interval, then \(f\) is a strictly increasing function. This is because when \(b > a\), \(f(b) - f(a) = f'(c) (b-a) > 0\). Likewise if \(f' < 0\) everywhere, then \(f\) is strictly decreasing.
Corollary (Generalized Mean Value Theorem)
Suppose \(f\) and \(g\) are both continuous, real valued functions on \([a,b]\) and differentiable on \((a,b)\). Then there exists \(c \in (a,b)\) such that
\[ (f(b) - f(a))g'(c) = (g(b) - g(a))f'(c). \]
Proof
Let
\[{}h(x){}\]
\[{}:= (f(b) - f(a)) g(x){}\]
\[{}- (g(b) - g(a)) f(x).{}\]
It is easy to check that
\[{}h(b) - h(a) {}\]
\[{}= (f(b) - f(a))(g(b)-g(a)){}\]
\[{}- (g(b)-g(a))(f(b)-f(a)){}\]
\[{}= 0.{}\]
This means that Rolle's Theorem implies that \(h'(c) = 0\) for some \(c \in (a,b)\). Recalling the definition of \(h\) gives the conclusion.

3. MVT Application: l'Hôpital's Rule

l'Hôpital's Rule is a name that covers a number of distinct situations in which a limit of a ratio can be computed by replacing both numerator and denominator with their derivatives. We will not prove all the various cases, but will give some representative proofs. In each case, the Generalized Mean Value Theorem is the idea that drives the argument.
Proposition
Suppose \(f\) and \(g\) are real, differentiable functions on some interval \(I \subset {\mathbb R}\) containing some point \(a\). Suppose also that \(f(a) = g(a) = 0\) and that neither \(g\) nor \(g'\) vanishes on \(I \setminus \{a\}\). Then for any \(x \in I \setminus \{a\}\), there is some point \(c \in I\) strictly between \(a\) and \(x\) such that
\[ \frac{f(x)}{g(x)} = \frac{f'(c)}{g'(c)}. \]
Proof
Meta (Main Idea)
This is a consequence of the Generalized Mean Value Theorem. If \(x > a\), the GMVT is applied on the interval \([a,x]\) to give some \(c \in (a,x)\) such that \(f(x) g'(c) = g(x) f'(c)\), which implies \(\frac{f(x)}{g(x)} = \frac{f'(c)}{g'(c)}\) because neither \(g(x)\) nor \(g'(c)\) vanishes. Note in particular that \(g\) doesn't vanish because \(g(a) = 0\) and \(g\) is strictly monotone on both sides of \(x=a\) because \(g' \neq 0\) there. If \(x < a\), the GMVT is instead applied to \([x,a]\) and proceeds analogously.

As a side remark, the Intermediate Value Property of derivatives implies that \(g'\) never equalling zero means that it's always positive or always negative. Applying the Mean Value Theorem itself then guarantees that \(g\) must be monotone. Thus if we assume that \(g' \neq 0\) on \(I \setminus \{a\}\), it's actually always going to be true that \(g \neq 0\) on \(I \setminus \{a\}\) as well (because already we know that \(g(a) = 0\) and \(g\) will be strictly monotone on both sides of \(a\)).
Corollary (l'Hôpital's Rule for \(\frac{0}{0}\) as \(x \rightarrow a\))
Suppose \(f\) and \(g\) are real, differentiable functions on some interval \(I \subset {\mathbb R}\) containing some point \(a\). Suppose also that \(f(a) = g(a) = 0\) and that neither \(g\) nor \(g'\) vanishes on \(I \setminus \{a\}\). If
\[ \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} = L \]
then
\[ \lim_{x \rightarrow a} \frac{f(x)}{g(x)} = L. \]
In other words, if the limit of the ratio \(f'/g'\) exists, then the limit of \(f/g\) does as well and the limits are equal.
Proof
By the proposition and the existence of the limit \(f'/g'\), for any \(\epsilon > 0\), there is a \(\delta > 0\) such that \(0 < |c-a| < \delta\) implies
\[ \left| \frac{f'(c)}{g'(c)} - L \right| < \epsilon. \]
If \(x\) is any point of the interval \(I\) such that \(0 < |x-a| < \delta\), then the corresponding point \(c\) such that
\[ \frac{f(x)}{g(x)} = \frac{f'(c)}{g'(c)}\]
will also have \(0 < |c-a| < \delta\) because \(c\) is between \(x\) and \(a\). Thus the equality of these two ratios gives
\[ \left| \frac{f(x)}{g(x)} - L \right| < \epsilon. \]

3.1. Some hints for other cases of l'Hôpital's Rule

  1. Type \(\frac{0}{0}\) as \(x \rightarrow \infty\): Assume that neither \(g\) nor \(g'\) vanishes for \(x\) sufficiently large. Then fix some large \(x \in {\mathbb R}\) and some \(y > x\) and use the Generalized Mean Value Theorem to prove that
    \[ \frac{f(y) - f(x)}{g(y)-g(x)} = \frac{f'(c)}{g'(c)}\]
    for some \(c\) between \(x\) and \(y\). The left-hand side can be made as close as desired to the ratio \(\frac{f(x)}{g(x)}\) by taking \(y\) sufficiently large because \(f(y),g(y) \rightarrow \infty\) as \(y \rightarrow \infty\). Thus \(f(x)/g(x)\) differs by at most \(\epsilon/2\) from \(f'(c)/g'(c)\) for some \(c > x\); this is enough to modify the argument above for this case.
  2. Type \(\frac{\infty}{\infty}\) as \(x \rightarrow a\): Here the trick is to argue based on the idea that subtracting off some finite amount from numerator and denominator is harmless. This idea alone is not quite enough, because we also need to guarantee that the point \(c\) at which \(f'(c)/g'(c)\) is evaluated can be forced to be as close as desired to \(a\). Assume that neither \(g\) nor \(g'\) vanishes for \(x\) near \(a\). For any distinct points \(x,y\) near to but not equal to \(a\),
    \[ \frac{f(y) - f(x)}{g(y)-g(x)} = \frac{f'(c)}{g'(c)}\]
    for some \(c\) between \(x\) and \(y\). If \(|f(y)|\) and \(|g(y)|\) are both larger than, for example, \(R \max \{|f(x)|,|g(x)|,1\}\), then
    \[ \frac{f(y) - f(x)}{g(y)-g(x)}\]
    differs from \(\frac{f(y)}{g(y)}\) by a factor somewhere between \(\frac{R-1}{R+1}\) and \(\frac{R+1}{R-1}\), so
    \[ \frac{R-1}{R+1} \frac{f'(c)}{g'(c)} \leq \frac{f(y)}{g(y)} \leq \frac{R+1}{R-1} \frac{f'(c)}{g'(c)}. \]
    One way to proceed is to choose \(\epsilon > 0\), let \(\delta\) be such that \(|f'(c)/g'(c) - L| < \epsilon/2\) when \(0 < |c-a| < \delta\); then fix \(x\) with \(0 < |x-a| < \delta\) and some \(R\) so large that
    \[ \frac{R+1}{R-1} \left(L + \frac{\epsilon}{2} \right) < L + \epsilon\]
    and
    \[ L - \epsilon < \frac{R-1}{R+1} \left(L - \frac{\epsilon}{2} \right)\]
    (which has to exist because the limits as \(R \rightarrow \infty\) are simply \(L + \frac{\epsilon}{2}\) and \(L - \frac{\epsilon}{2}\), respectively). Now Consider any \(y\) sufficiently close to \(a\) so that
    \[{}|f(y)|,|g(y)| {}\]
    \[{}\geq R \max \{|f(x)|,|g(x)|,1\},{}\]
    it will follow that
    \[{}L - \epsilon{}\]
    \[{}< \frac{R-1}{R+1} \frac{f'(c)}{g'(c)}{}\]
    \[{}\leq \frac{f(y)}{g(y)}{}\]
    \[{}\leq \frac{R+1}{R-1} \frac{f'(c)}{g'(c)}{}\]
    \[{}< L + \epsilon.{}\]
  3. Type \(\frac{\infty}{\infty}\) as \(x \rightarrow \infty\): Can be handled by essentially the same argument as above, but taking \(x\) and \(y\) to be sufficiently large instead of sufficiently close to \(a\).