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प्रश्न
f(x) = x(x – 1)2 in [0, 1]
योग
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उत्तर
We have, f(x) = x(x – 1)2 in [0, 1]
Since, f(x) = x(x – 1)2 is a polynomial function it is continuous in [0,1] and differentiable in (0, 1)
Now, f(0) = 0 and f(1)
⇒ f(0) = f(1)
f satisfies the conditions of Rolle's theorem.
Hence, by Rolle's theorem there exists atleast one c ∈ (0, 1) such that f'(c) = 0
⇒ 3c2 – 4c + 1 = 0
⇒ (3c – 1)(c – 1) = 0
⇒ c = `1/3 ∈ (0, 1)`
Therefore, Rolle's theorem has been verified.
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