Mean Value Theorem

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The mean value theorem (MVT) says that, for a given arc connecting two points of a function, there is at least one point at which the slope of the tangent line is equal to the slope of the arc.



Theorem (Rolle's theorem)

Let \(f(x)\) be a continuous function defined on a closed interval \([a, b]\). Suppose \(f(x)\) is differentiable on the open interval \((a, b)\) and \(f(a) = f(b)\). Then there exists a \(c\in (a,b)\) such that \(f'(c) = 0\).

Proof. If the (global) maximum and minimum are both at \(x=a\) and \(x=b\), then \(f(x)\) is constant since \(f(a) = f(b)\) (= maximum = minimum). In this case, \(f'(x)=0\) at all \(x \in (a,b)\). Otherwise, \(f(x)\) has a maximum or minimum value at some \(c \in (a,b)\) so that \(f'(c) = 0\). ■

Theorem (Mean value theorem)

Let \(f(x)\) be a continuous function on \([a,b]\) such that it is differentiable on \((a,b)\). Then there exists a \(c \in (a,b)\) such that

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

Proof. Let us define

\[g(x) = f(x) - \frac{f(b) - f(a)}{b - a}\cdot(x - a).\]

Then, \(g(x)\) is continuous on \([a, b]\) and differentiable on \((a,b)\), and \(g(a) = g(b) = f(a)\). Hence, by Rolle's theorem, there exists a \(c\in(a,b)\) such that \(g'(c) = 0\). But

\[g'(c) = f'(c) - \frac{f(b) - f(a)}{b - a} = 0\]

so

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

■

Let's see some applications of the mean value theorem.

Corollary

Suppose that the function \(f(x)\) is differentiable on the open interval \(I=(a,b)\) and identically \(f'(x) = 0\) on \(I\). Then \(f(x)\) is a constant function on \(I\). That is, \(f(x) = C\) for all \(x \in I\) where \(C\) is a constant.

Proof. Let us pick an arbitrary \(c\in (a,b)\) and let \(C = f(c)\). We show that for all \(x\in(a,b)\), \(f(x) = C\). If \(x=c\), then \(f(x) = C\) by definition of \(C\). Suppose \(x\neq c\). If \(x < c\), consider the closed interval \([x, c]\). Since \(f(x)\) is differentiable on \((a,b)\), it is continuous on \([x, c]\). By the mean value theorem, there is a \(d\in(x,c)\) such that \(f(x) - f(c) = f'(d)(x-c)\). But since \(f'(d) = 0\) (by assumption), \(f(x) = f(c) = C\). If \(x > c\), we consider the closed interval \([c, x]\) and apply a similar argument. ■

Corollary (cor:monotone)

Suppose the function \(f(x)\) is continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\).

  1. If \(f'(x) > 0\) for all \(x\in (a,b)\), then \(f(x)\) is strictly monotone increasing on \([a,b]\).
  2. If \(f'(x) < 0\) for all \(x\in (a,b)\), then \(f(x)\) is strictly monotone decreasing on \([a,b]\).
Proof. We only show part 1. Part 2 is similar.
It suffices to show that, for any \(c, d\in [a,b]\), if \(c < d\) then \(f(c) < f(d)\).
By assumption, \(f(x)\) is continuous on \([c,d]\) and differentiable on \((c,d)\). By the mean value theorem, there exists a \(t\in(c,d)\) such that \(f(d) - f(c) = f'(t)(d - c)\). But since \(f'(t) > 0\), we have \(f'(t)(d - c) > 0\) so that \(f(d) - f(c) > 0\). ■

Corollary (cor:minmax)

Suppose that the function \(f(x)\) is twice differentiable and that \(f'(c) = 0\) at some \(c\in (a,b)\).
  1. If \(f''(c) > 0\), then \(x = c\) is a local minimum point.
  2. If \(f''(c) < 0\), then \(x = c\) is a local maximum point.
Proof. We prove only part 1 (Part 2 is similar).
By assumption,
\[\lim_{x\to c}\frac{f'(x) - f'(c)}{x-c} = \lim_{x\to c}\frac{f'(x)}{x - c} = f''(c) > 0.\]
Let \(\varepsilon = f''(c) (> 0)\). There exists a \(\delta > 0\) such that, for all \(x \in (a,b)\),  \(0 < |x - c| < \delta\) implies \(\left|\frac{f'(x)}{x - c} - f''(c)\right| < \varepsilon\) or
\[-\varepsilon < \frac{f'(x)}{x - c} - f''(c) < \varepsilon.\]
It follows that
\[\frac{f'(x)}{x - c} > f''(c) -\varepsilon = 0.\]
In particular, if \(c - \delta < x < c\), then \(x - c < 0\) so \(f'(x) < 0\). That is, for all \(x \in (c -\delta, c)\), \(f(x) > f(c)\) (Corollary {cor:monotone}).

If \(c < x < c + \delta\), we have \(f'(x) > 0\) so, by the same argument (Corollary {cor:monotone}), \(f(c) < f(x)\).

In summary, for all \(x \in (c - \delta, c + \delta)\), if \(x \neq c\), then \(f(x) > f(c)\), which shows that \(f(c)\) is a local minimum value. ■

Definition (Convex, concave)

Let \(f(x)\) be a function defined on an open interval \(I\). \(f(x)\) is said to be convex at \(x= c\) if the graph of \(y= f(x)\) is above its tangent line at \(x=c\) in the neighbor of \(x=c\), that is, if there exists a \(\delta > 0\) such that, for all \(x\in I\),  \(0 < |x - c| < \delta\) implies \(f(x) > f'(c)(x - c) + f(c)\).

Similarly, \(f(x)\) is said to be concave at \(x = c\) if the graph of \(y = f(x)\) is below its tangent line at \(x=c\) in the neighbor of \(x=c\), that is, there exists a \(\delta > 0\) such that, for all \(x \in I\), \(0 < |x - c| < \delta\) implies \(f(x) < f'(c)(x - c) + f(c)\).

Remark. Recall that \(y = f'(c)(x - c) + f(c)\) is the equation of the tangent line of \(f(x)\) at \(x=c\). □

Corollary 

Let \(f(x)\) be a twice differentiable function on an open interval \(I\) and \(c\in I\).
  1. If \(f''(c) > 0\), then \(f(x)\) is convex at \(x=c\). 
  2. If \(f''(c) < 0\), then \(f(x)\) is concave at \(x=c\).
Proof. We prove only part 1. Part 2 is similar.

Let us define \(g(x) = f(x) - [f'(c)(x - c) + f(c)]\). \(g(x)\) is twice differentiable on \(I\) and \(g'(x) = f'(x) - f'(c)\) so \(g(c) = g'(c) = 0\). Also, \(g''(x) = f''(x)\) so that \(g''(c) = f''(c) > 0\). Therefore, by Corollary {cor:minmax}, \(x=c\) is a local minimum point of \(g(x)\). By the definition of local minimum, it follows that \(f(x)\) is convex \(x = c\). ■

Example. Consider \(f(x) = e^{-x^2}\). Let us find the range of \(x\) where \(f(x)\) is convex and where it is concave. Let us also find the extrema of \(f(x)\).
\[ \begin{eqnarray} f'(x) &=& -2xe^{-x^2},\\ f''(x) &=& -2e^{-x^2} + 4x^2e^{-x^2} = 2(2x^2 - 1)e^{-x^2}. \end{eqnarray} \]
Solving \(f''(x) = 0\) yields \(x = \pm\frac{1}{\sqrt{2}}\). Therefore,
  • \(f''(x) < 0\) if \(-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}\) where \(f(x)\) is concave, and
  • \(f''(x) > 0\) if \(x < -\frac{1}{\sqrt{2}}\) or \(x > \frac{1}{\sqrt{2}}\) where \(f(x)\) is convex.
\(f'(x) = 0\) only if \(x = 0\) and \(f''(0) = -2 < 0\). So \(f(0)\) is a local maximum value (in fact, it is the global maximum value). □

As in the above example, the point at which the sign of the second-derivative changes is called an inflection point.

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