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Q.Verify Rolle's theorem for the function x2−1x^2 - 1 on [−1,1][-1, 1].

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2025Subjective· 2mImportance★★★★★
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f(x)=x2−1f(x)=x^2-1 satisfies all three conditions of Rolle's theorem on [−1,1][-1,1], and the required point c=0c=0 exists in the open interval.

Rolle's theorem requires: (i) ff continuous on [a,b][a,b], (ii) ff differentiable on (a,b)(a,b), (iii) f(a)=f(b)f(a)=f(b). If all hold, there exists c∈(a,b)c\in(a,b) with f′(c)=0f'(c)=0.

Check conditions:

f(x)=x2−1f(x)=x^2-1 is a polynomial, hence continuous on [−1,1][-1,1] and differentiable on (−1,1)(-1,1).

f(−1)=(−1)2−1=0f(-1) = (-1)^2-1 = 0

f(1)=(1)2−1=0f(1) = (1)^2-1 = 0

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