Lagrange’s Theorem
Statement
The Lagrange’s theorem, also known as the mean value theorem, states the following. Consider a function $f ( x )$, continuous in the closed and bounded interval $[ a , b ]$ and differentiable at every point inside the interval. Then, there exists at least one point $c$ inside the interval such that the following relation holds:
\[f^{'} ( c ) = \frac{f ( b ) - f ( a )}{b - a}\]This means that there exists at least one point where the derivative of the function is equal to the slope of the secant line connecting $a$ and $b$. In other words, at some point in the interval, the instantaneous rate of change of the function matches its average rate of change.
A geometric view of Lagrange’s theorem
From a geometric point of view, the theorem states that there exists at least one point $c$ where the tangent line at that point is parallel to the secant line connecting points $A$ and $B$ on the graph.
In the right-angled triangle $A B H$, we have $\overset{―}{B H} = \overset{―}{A H} \cdot tan \alpha$ that is:
\[tan \alpha = \frac{\overset{―}{B H}}{\overset{―}{A H}}\]We have:
\[\overset{―}{B H} = f ( b ) - f ( a ) , \overset{―}{A H} = b - a\]The slope of segment $A B$ is equal to $tan \alpha$ that is:
\[tan \alpha = \frac{f ( b ) - f ( a )}{b - a}\]Since the tangent at $c$ to the curve is parallel to $A B$, both have the same slope, hence:
\[f^{'} ( c ) = \frac{f ( b ) - f ( a )}{b - a}\]Proof
To prove Lagrange’s theorem we define an auxiliary function $\varphi ( x )$ as follows:
\[\varphi ( x ) = f ( x ) - f ( a ) - \frac{f ( b ) - f ( a )}{b - a} \cdot ( x - a )\]We verify that $\varphi ( x )$ satisfies the hypotheses of Rolle’s Theorem:
- $f ( x )$ is continuous on the closed interval $[ a , b ]$.
- $f ( x )$ is differentiable on the open interval $( a , b )$.
- $\varphi ( a ) = \varphi ( b )$.
By calculating $\varphi ( a )$ and $\varphi ( b )$, we obtain:
\[\varphi ( a ) = f ( a ) - f ( a ) - \frac{f ( b ) - f ( a )}{b - a} \cdot ( a - a ) = 0\] \[\varphi ( b ) & = f ( b ) - f ( a ) - \frac{f ( b ) - f ( a )}{b - a} \cdot ( b - a ) \\ & = f ( b ) - ( f ( b ) - f ( a ) ) = 0\]Therefore, $\varphi ( a ) = \varphi ( b ) = 0$.
Applying Rolle’s Theorem to $\varphi ( x )$, we find that there exists at least one point $c$ within the interval $] a , b [$ such that $\varphi^{‘} ( c ) = 0$. Calculating the derivative of $\varphi ( x )$, we have:
\[\varphi^{'} ( x ) = f^{'} ( x ) - \frac{f ( b ) - f ( a )}{b - a}\]Now, we calculate the derivative of $\varphi ( x )$ at the point $c$ and set it equal to 0. From this, we obtain:
\[\varphi^{'} ( c ) = f^{'} ( c ) - \frac{f ( b ) - f ( a )}{b - a} = 0\]That is:
\[f^{'} ( c ) = \frac{f ( b ) - f ( a )}{b - a}\]which corresponds exactly to the thesis we wanted to prove.
A special case: when the derivative is zero everywhere
From Lagrange’s theorem, it follows that if a function $f ( x )$ is continuous on the interval $[ a , b ]$, differentiable on the interval $] a , b [$, and $f^{‘} ( x )$ is zero at every point in the interior of the interval, then $f ( x )$ is constant on the entire interval $[ a , b ]$.
Indeed, if we take a point $\overset{―}{x} \in [ a , b ]$ and apply the theorem to the interval $[ a , \overset{―}{x} ]$, then there exists a point $c \in ] a , \overset{―}{x} [$ such that: \(f^{'} ( c ) = \frac{f ( \overset{―}{x} ) - f ( a )}{\overset{―}{x} - a}\)
Since $f^{‘} ( \overset{―}{x} ) = 0$ for every point in $] a , b [$, it follows that $f^{‘} ( c ) = 0$ as well. For this reason, it must be:
\[f ( \overset{―}{x} ) – f ( a ) = 0 \rightarrow f ( \overset{―}{x} ) = f ( a ) \forall \overset{―}{x} \in [ a , b ]\]For this reason, $f$ is constant on the entire interval $[ a , b ]$.
A numerical example
To see the theorem at work, let us examine a concrete case. We want to identify a point $c$ inside a given interval where the instantaneous rate of change of a function coincides with its average rate of change on that interval. The computation also shows that such a point is not something one can usually predict by inspection. Consider the polynomial:
\[f ( x ) = x^{3} - 4 x^{2} + x + 6\]on the closed interval $[ 1 , 4 ]$. Being a polynomial, $f$ is continuous on $[ 1 , 4 ]$ and differentiable on $( 1 , 4 )$. The assumptions of Lagrange’s theorem are therefore satisfied. We begin by evaluating the function at the endpoints:
\(f ( 1 ) = 1 - 4 + 1 + 6 = 4\) \(f ( 4 ) = 64 - 64 + 4 + 6 = 10\)
The slope of the secant line through the points $( 1 , 4 )$ and $( 4 , 10 )$ is
\[\frac{f ( 4 ) - f ( 1 )}{4 - 1} = \frac{10 - 4}{3} = 2\]This number represents the average rate of change of $f$ over $[ 1 , 4 ]$. Next we compute the derivative:
\[f^{'} ( x ) = 3 x^{2} - 8 x + 1\]We look for values of $c$ such that $f^{‘} ( c ) = 2.$ This leads to the equation:
\(3 c^{2} - 8 c + 1 = 2\) \(3 c^{2} - 8 c - 1 = 0\)
Applying the quadratic formula gives:
\[c = \frac{8 \pm \sqrt{64 + 12}}{6} = \frac{8 \pm \sqrt{76}}{6} = \frac{4 \pm \sqrt{19}}{3}\]We obtain two candidates:
\(c_{1} = \frac{4 - \sqrt{19}}{3} \approx 0.21\) \(c_{2} = \frac{4 + \sqrt{19}}{3} \approx 2.79\)
Lagrange’s theorem guarantees a solution inside the open interval $] 1 , 4 [$. Among the two values found, only:
\[c = \frac{4 + \sqrt{19}}{3} \approx 2.79\]lies in $] 1 , 4 [$. The other root falls outside the interval and is therefore not relevant in this context. At this point, the tangent line to the graph of $f$ is parallel to the secant line joining $( 1 , 48 )$ and $( 4 , 10 )$.
This example shows that the equation $f^{‘} ( c ) = 2$ may admit more than one solution, yet only those inside the interval are meaningful for the theorem.
The solution is: \(c = \frac{4 + \sqrt{19}}{3}\)
Selected references
- Harvard University, O. Knill. The Mean Value Theorem
- Stony Brook University. Lecture 18: Mean Value Theorem
- University of California Davis, D. Kouba. Mean Value Theorem – Problems and Proofs
- University of Cambridge, W. T. Gowers. What is the point of the Mean Value Theorem?