Let f satisfy the conditions of Lagrange's theorem

Let $f$ satisfy the conditions of Lagrange's theorem on $[a,b]$, and set \[ F(x) = \begin{vmatrix} x & f(x) & 1 \\ b & f(b) & 1 \\ a & f(a) & 1 \end{vmatrix}. \] Prove Lagrange's theorem by applying Rolle's theorem to $F$, and explain the geometric meaning of $F$.

Step 1: Expand along the first row

The three minors are

\[ \begin{vmatrix} f(b) & 1 \\ f(a) & 1 \end{vmatrix} = f(b)-f(a), \qquad \begin{vmatrix} b & 1 \\ a & 1 \end{vmatrix} = b-a, \qquad \begin{vmatrix} b & f(b) \\ a & f(a) \end{vmatrix} = b\,f(a)-a\,f(b), \]

so that

\[ F(x) = x\bigl[f(b)-f(a)\bigr]-f(x)\bigl(b-a\bigr)+b\,f(a)-a\,f(b). \tag{1} \]

Only the term $-f(x)\bigl(b-a\bigr)$ is not linear in $x$; everything else is a constant or a multiple of $x$.

Step 2: Rolle's hypotheses for $F$

By (1), $F$ inherits continuity on $[a,b]$ and differentiability on $(a,b)$ directly from $f$. For the endpoint values, return to the determinant: \[F(a) = \begin{vmatrix} a & f(a) & 1 \\ b & f(b) & 1 \\ a & f(a) & 1 \end{vmatrix} = 0, \qquad F(b) = \begin{vmatrix} b & f(b) & 1 \\ b & f(b) & 1 \\ a & f(a) & 1 \end{vmatrix} = 0,\]

each vanishing because two rows coincide. So $F(a) = F(b) = 0$.

Step 3: Apply Rolle's theorem

There exists $\xi\in(a,b)$ with $F'(\xi) = 0$. Differentiating (1), $F'(x) = \bigl[f(b)-f(a)\bigr]-f'(x)\bigl(b-a\bigr)$.

\[ F'(\xi) = 0 \;\Longrightarrow\; f'(\xi) = \frac{f(b)-f(a)}{b-a}, \]

which is Lagrange's theorem.

Step 4: What $F$ means

The signed area of a triangle with vertices $\bigl(x_{1},y_{1}\bigr),\bigl(x_{2},y_{2}\bigr),\bigl(x_{3},y_{3}\bigr)$ is half the determinant of the same shape. Hence

\[ F(x) = 2\times\bigl(\text{signed area of the triangle } M A B\bigr), \qquad M = \bigl(x,f(x)\bigr),\ A = \bigl(a,f(a)\bigr),\ B = \bigl(b,f(b)\bigr). \]

The two conditions $F(a) = F(b) = 0$ then say something obvious: when $M$ coincides with $A$ or with $B$ the triangle degenerates and its area is zero. Rolle's point $\xi$ is where this area is stationary — that is, where $M$ is farthest from the chord, and the tangent there is parallel to $AB$.

Step 5: Verification

\[f(x) = x^{2},\ a = 0,\ b = 2:\ F(x) = 4x-2x^{2}+0 = 2x\bigl(2-x\bigr)\]

\[F(0) = F(2) = 0 \;\;\checkmark, \qquad F'(x) = 4-4x = 0 \;\Longrightarrow\; \xi = 1\]

\[f'(1) = 2 = \frac{f(2)-f(0)}{2-0} = \frac{4}{2} \;\;\checkmark\]