If f(x) and g(x) are differentiable functions for 0 ≤ x ≤ 23 such that f(0) = 2, g(0) = 0, f(23) = 22, g(23) = 10, then show that f'(x) = 2g'(x) for at least one x in the interval (0, 23).
Text Solution
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Let φ (x) = f(x) – 2g(x) ; x ∈ [0, 23]
⇒ φ′ (x) = f'(x) – 2g ′ (x)
Also φ (0) = f(0) – 2g(0) = 2 – 0 = 2
φ (23) = f(23) – 2g(23) = 22 – 20 = 2
Since f(x) and g(x) are differentiable in [0, 23] hence φ (x) is also continous in [0, 23] and differentiable in [0, 23], so all the conditions of Rolle's theorem are satisfied. Hence there exist a number c, 0 < c < 23 for which φ '(x) = 0
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