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If f:[– 5, 5] → R is differentiable and if f’(x) doesn’t vanish anywhere, then prove that f(– 5) ≠ f(5). |
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Answer» Given that f is continuous and differentiable in the interval [– 5, 5]. It is also given that f’(x) doesn’t vanish anywhere. According to Rolle’s theorem for a differentiable function on [a, b] will have atleast one c∈(a, b) such that f’(c) = 0, if the following condition had satisfied: ⇒ f(a) = f(b). According to the problem it is given for any value of x, say r the values never equals to zero. ⇒ f’(r) ≠ 0 This is possible when Rolle’s theorem is not applicable. Let us Recap the Rolle’s theorem: For a Real valued function ‘f’: a) The function ‘f’ needs to be continuous in the closed interval [a, b]. b) The function ‘f’ needs differentiable on the open interval (a, b). c) f(a) = f(b) Then there exists at least one c in the open interval (a, b) such that f’(c) = 0. First, two conditions are satisfied according to the problem, so the only condition that cannot be satisfied is (c). So, we can clearly say that f(– 5) ≠ f(5). |
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