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Q.Verify Lagrange's mean value theorem for f(x)=x3−2x2−x+3f(x) = x^3 - 2x^2 - x + 3 on [1,2][1, 2].

Odisha ChseOdisha CHSE +2 Science Board Exam 2019Subjective· 4mImportance★★★★★
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ff is a polynomial (continuous and differentiable everywhere), so LMVT applies on [1,2][1,2]; solving f′(c)=f(2)−f(1)2−1f'(c)=\dfrac{f(2)-f(1)}{2-1} gives c=2+73∈(1,2)c=\dfrac{2+\sqrt7}{3}\in(1,2).

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

Being a polynomial, ff is continuous on [1,2][1,2] and differentiable on (1,2)(1,2), so Lagrange's Mean Value Theorem (LMVT) applies: there exists c∈(1,2)c\in(1,2) such that

f′(c)=f(2)−f(1)2−1f'(c) = \dfrac{f(2)-f(1)}{2-1}

Compute endpoint values:

f(1)=1−2−1+3=1f(1) = 1-2-1+3 = 1

f(2)=8−8−2+3=1f(2) = 8-8-2+3 = 1

f(2)−f(1)2−1=1−11=0\dfrac{f(2)-f(1)}{2-1} = \dfrac{1-1}{1} = 0

Compute f′(x)f'(x):

f′(x)=3x2−4x−1f'(x) = 3x^2-4x-1

Solve f′(c)=0f'(c)=0:

3c2−4c−1=0  ⟹  c=4±16+126=4±286=4±276=2±733c^2-4c-1=0 \implies c = \dfrac{4\pm\sqrt{16+12}}{6} = \dfrac{4\pm\sqrt{28}}{6} = \dfrac{4\pm2\sqrt7}{6} = \dfrac{2\pm\sqrt7}{3}

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