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Q.Verify Rolle's Theorem for the function f(x)=x2+2x−8f(x) = x^2 + 2x - 8, x∈[−4,2]x \in [-4, 2].

Manipur CohsemCOHSEM Manipur Higher Secondary Board 2023Subjective· 4mImportance★★★★★
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Check the three Rolle's Theorem conditions (continuity, differentiability, equal endpoint values), then explicitly find cc with f′(c)=0f'(c)=0.

f(x)=x2+2x−8f(x) = x^2+2x-8 on [−4,2][-4,2].

Condition 1 (Continuity): ff is a polynomial, hence continuous on [−4,2][-4,2]. ✓

Condition 2 (Differentiability): ff is a polynomial, hence differentiable on (−4,2)(-4,2). ✓

Condition 3 (f(−4)=f(2)f(-4)=f(2)):

f(−4)=(−4)2+2(−4)−8=16−8−8=0f(-4) = (-4)^2+2(-4)-8 = 16-8-8 = 0

f(2)=22+2(2)−8=4+4−8=0f(2) = 2^2+2(2)-8 = 4+4-8 = 0

So f(−4)=f(2)=0f(-4)=f(2)=0. ✓

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