Lagrange’s Theorem

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Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/lagrange-theorem/ Fetched from algebrica.org post 6193; source modified 2026-02-25T22:38:28.

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’ \left (c \right ) = \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.

Lagrange’s theorem.

In the right-angled triangle $ABH$, we have $\overline{BH} = \overline{AH} \cdot \tan{\alpha}$ that is:

\tan{\alpha} = \frac{\overline{BH}}{\overline{AH}}

We have:

\overline{BH} = f(b)-f(a), \quad \overline{AH} = b-a

The slope of segment $AB$ 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 $AB$, both have the same slope, hence:

f’ \left(c \right) = \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
\begin{align} \varphi(b) &= f(b)-f(a)-\frac{f(b)-f(a)}{b-a} \cdot (b-a)\\[0.5em] &= f(b)-(f(b)-f(a)) = 0 \end{align}

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 $\varphi’ \left(c \right) = 0$. 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’ \left(c \right) = f’ \left(c \right)-\frac{f(b)-f(a)}{b-a} = 0

That is:

f’ \left(c \right)=\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 $\overline{x} \in [a,b]$ and apply the theorem to the interval $[a, \overline{x}]$, then there exists a point $c \in ]a, \overline{x}[$ such that:

f’\left( c \right) = \frac{f(\overline{x})-f(a)}{\overline{x}-a}

Since $f’(\overline{x}) = 0$ for every point in $]a,b[$, it follows that $f’\left( c \right) = 0$ as well. For this reason, it must be:

f(\overline{x}) – f(a) = 0 \rightarrow f(\overline{x}) = f(a) \, \forall \, \overline{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 - 4x^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) = 3x^2 - 8x + 1

We look for values of $c$ such that $f’(c) = 2.$ This leads to the equation:

3c^2 - 8c + 1 = 2
3c^2 - 8c - 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 = \dfrac{4+\sqrt{19}}{3}

Selected references