Rolle’s Theorem Example – Quadratic on [-3,4]

Problem Statement

• Example: Verify Rolle's Theorem for $f(x)=x^2-x-12$ on the closed interval $[-3,4]$.
• Tasks:
  • Show all 3 hypotheses of Rolle’s Theorem are satisfied.
  • Identify the $c$ value(s) in the open interval $(-3,4)$ for which $f'(c)=0$.

Rolle’s Theorem ‑ Key Requirements

• Closed interval endpoints labeled $a=-3$ and $b=4$.
• Necessary hypotheses:
  1. $f$ is continuous on $[a,b]$.
  2. $f$ is differentiable on $(a,b)$.
  3. $f(a)=f(b)$.
• Guarantee: If 1–3 hold, then $\exists\,c\in(a,b)$ such that $f'(c)=0$.

Verifying $f(a)=f(b)$

• Evaluate at $x=-3$:
  • $f(-3)=(-3)^2-(-3)-12=9+3-12=0$.
• Evaluate at $x=4$:
  • $f(4)=4^2-4-12=16-4-12=0$.
• Result: $f(-3)=f(4)=0$ ⟹ third hypothesis satisfied.

Continuity & Differentiability Check

• $f$ is a polynomial.
• Polynomials are:
  • Continuous $\forall x\in\mathbb{R}$.
  • Differentiable $\forall x\in\mathbb{R}$.
• Thus hypotheses 1 and 2 are automatically met on $[-3,4]$.

Finding $c$ such that $f'(c)=0$

• Compute derivative:
  • $f'(x)=2x-1$.
• Solve $f'(x)=0$:
  • $2x-1=0 \Rightarrow 2x=1 \Rightarrow x=\tfrac12$.
• Interval membership:
  • $\tfrac12\in(-3,4)$ ⟹ valid $c$.
• No other zeros because $f'(x)$ is linear (degree 1).

Conclusion

• All Rolle’s Theorem conditions satisfied for $f(x)=x^2-x-12$ on $[-3,4]$.
• The theorem guarantees (at least) one root of the derivative in $(a,b)$; we found
  • $c=\tfrac12$.
• Because the derivative is linear, this is the unique $c$.

Extra Insights & Connections

• Visual interpretation:
  • Since $f(-3)=f(4)=0$, graph touches the $x$-axis at both ends.
  • Rolle’s Theorem predicts a horizontal tangent (slope $=0$) somewhere between; occurs at $x=0.5$.
• Relation to Mean Value Theorem (MVT):
  • Rolle’s Theorem is a special case of the MVT where $f(a)=f(b)$, letting the MVT slope $\big(f(b)-f(a)\big)/(b-a)=0$.
• Practical implication: Confirms existence of stationary point between two real roots of a quadratic.