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Question:

If f and g are differentiable functions in [0, 1] satisfying f(0) = 2 = g(1), g(0) = 0 and f(1) = 6, then for some c ∈ [0, 1]

2f'(c)=g'(c)

f'(c)=g'(c)

2f'(c)=3g'(c)

f'(c)=2g'(c)

Solution:

Let h(x) = f(x) - 2g(x) as h(0) = h(1) = 2.
Hence, using Rolle's theorem h'(c) = 0 ⇒ f'(c) = 2g'(c)