Rolle's Theorem or Rolle's Lemma

Rolle's Theorem (or Rolle's Lemma): if a real-valued function $f$ is continuous on the closed interval $[a,b]$, differentiable on the open interval $(a,b)$, and $f(a)=f(b)$, then there exists at least one point $c$ in the open interval $(a,b)$ such that $f'(c)=0$.

In words: any real-valued, differentiable function that takes the same value at two distinct points must have at least one stationary point (where the first derivative — the tangent's slope — is zero) somewhere strictly between them.

Geometrical significance. If $f(x)$ is continuous on $[a,b]$ (so its graph can be drawn without lifting the pen from $x=a$ to $x=b$) and differentiable on $(a,b)$ (so the graph has a well-defined tangent at every interior point, with no sharp corners or vertical tangents), and if the graph starts and ends at the same height ($f(a)=f(b)$), then the graph must rise and then fall (or fall and then rise) somewhere in between — and at the very top (or bottom) of that rise-and-fall, the tangent to the curve is momentarily horizontal, i.e. parallel to the X-axis, since $f(a)=f(b)$ forces the curve to "return" to its starting height.

Worked Example 1 — Checking the hypotheses. (i) For $f(x)=2x^3-5x^2+3x+2$ on $\left[0,\dfrac32\right]$: this is a polynomial, so it is automatically continuous on the closed interval and differentiable on the open interval. Checking the endpoint values: $f(0)=2$ and $f\left(\dfrac32\right)=2\left(\dfrac{27}{8}\right)-5\left(\dfrac94\right)+3\left(\dfrac32\right)+2=\dfrac{54-90+36}{8}+2=2$. Since $f(0)=f\left(\dfrac32\right)=2$, all the conditions of Rolle's theorem are satisfied. (ii) For $f(x)=x^2-2x+3$ on $[1,4]$: again a polynomial (smooth everywhere), but $f(1)=2$ while $f(4)=11$; since $f(1)\ne f(4)$, the conditions of Rolle's theorem are NOT satisfied for this function on this interval.

Worked Example 2 — Full verification, finding $c$. Verify Rolle's theorem for $f(x)=x^2-4x+10$ on $[0,4]$. As a polynomial, $f$ is continuous and differentiable everywhere. $f(0)=10$ and $f(4)=16-16+10=10$, so $f(0)=f(4)=10$: all conditions hold. Differentiating: $f'(x)=2x-4=2(x-2)$. Setting $f'(c)=0$ gives $c=2$, and indeed $2\in(0,4)$ — Rolle's theorem is verified.

Worked Example 3 — Finding $b$ given $a$. Given that Rolle's theorem holds for $f(x)=x^3-2x^2+3$ on some interval $[a,b]$ with $a=0$, find $b$. Write $f(x)=g(x)+3$ where $g(x)=x^3-2x^2=x^2(x-2)$, which is zero exactly at $x=0$ and $x=2$. So $f(0)=g(0)+3=3$ and $f(2)=g(2)+3=3$, i.e. $f(0)=f(2)=3$. Since $f$ is a cubic polynomial (continuous and differentiable everywhere), all the conditions of Rolle's theorem are satisfied on $[0,2]$, so $b=2$. …