Prove the following variant Theorem 4.14. Suppose f [a, b] → R is -1,1. If f is differentiable at c = [a, b], f'(c) ‡ 0, and f-¹ is continuous at d = f(c), then f-¹ is differentiable at d and (f-¹')'(d) = = 1 f'(c)

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Q: {6} = I ( (b) T = [4,9) %3D (c) T = [-4,9] -

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