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Verify Rolle's theorem: x(x+3)/e^(x/2)

Short answer

Rolle's theorem: if f is continuous on [a,b], differentiable on (a,b), and f(a)=f(b), then ∃c∈(a,b) with f'(c)=0. For f(x)=x(x+3)e^{-x/2} on [-3,0], f(-3)=f(0)=0, and f'(c)=0 gives c=−2 in (-3,0), verifying the theorem.

State Rolle's Theorem and verify the theorem for f(x) = x(x+3)/e^(x/2); x ∈ [−3, 0].

Rolle's theorem

If f is continuous on [a, b], differentiable on (a, b) and f(a) = f(b), then there is at least one c in (a, b) such that f′(c) = 0.

Verification

f(x) = x(x + 3) e^(−x/2) on [−3, 0].

  1. Continuity and differentiability: f is a product of a polynomial and an exponential, so it is continuous on [−3, 0] and differentiable on (−3, 0).
  2. Equal end values: f(−3) = (−3)(0)e^(3/2) = 0 and f(0) = 0·3·1 = 0. So f(−3) = f(0).
  3. Derivative:
f′(x) = e^(−x/2) [ (2x + 3) − x(x+3)/2 ]
      = e^(−x/2) (−x² + x + 6)/2
      = −(1/2)(x − 3)(x + 2) e^(−x/2)

Setting f′(c) = 0 gives c = 3 or c = −2. Only c = −2 lies in (−3, 0).

Hence Rolle's theorem is verified with c = −2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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