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Rolle vs Lagrange; verify Lagrange's MVT

Short answer

Rolle's theorem requires f(a)=f(b) and gives f'(c)=0; Lagrange's MVT does not require equal endpoints and yields f'(c) = (f(b)−f(a))/(b−a). For f(x)=x(x−1)(x−2) on [0,1/2], f(0)=0, f(1/2)=3/8, slope 3/4. f'(x)=3x²−6x+2; solving f'(c)=3/4 gives c = 1 − √21/6 ≈ 0.236 in (0,1/2), verifying the theorem.

State difference between Rolle's Theorem and Lagrange's Mean value theorem. Verify Lagrange's mean value theorem for f(x) = x(x−1)(x−2) when x ∈ [0, 1/2].

Difference between Rolle's theorem and Lagrange's mean value theorem

PointRolle's theoremLagrange's MVT
Conditionsf continuous on [a,b], differentiable on (a,b), and f(a) = f(b)Same, but f(a) = f(b) is not needed
Conclusionf′(c) = 0 for some c in (a,b)f′(c) = (f(b) − f(a))/(b − a)
GeometryTangent parallel to the x-axisTangent parallel to the chord joining the end points
RelationA special case of Lagrange's MVTThe general form (Rolle is the case f(a) = f(b))

Verification of Lagrange's MVT

f(x) = x(x − 1)(x − 2) = x³ − 3x² + 2x on [0, 1/2].

  1. f is a polynomial, hence continuous on [0, 1/2] and differentiable on (0, 1/2).
  2. f(0) = 0 and f(1/2) = (1/2)(−1/2)(−3/2) = 3/8.
  3. Slope of chord = (3/8 − 0)/(1/2 − 0) = 3/4.
  4. f′(x) = 3x² − 6x + 2. Set f′(c) = 3/4:

3c² − 6c + 5/4 = 0, so 12c² − 24c + 5 = 0, giving c = [24 ± √(576 − 240)]/24 = 1 ± √21/6.

c = 1 + √21/6 ≈ 1.764 (rejected, outside the interval); c = 1 − √21/6 ≈ 0.236, which lies in (0, 1/2).

Hence Lagrange's MVT is verified with c = 1 − √21/6 ≈ 0.236.

✓ Answer checked by Claude Opus 5.5 (high effort)

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