Let I be a neighborhood of x_0 (that's x subscript 0) and let f: I->R be continuous, strictly monotone, and differentiable at x_0. Assume that f '(x_0)=0. Use the property: f^(-1)(f(x))=x (that's the inverse function of the function f(x) = x) for x in I and the chain rule to prove that the inverse function f^(-1): f(I)->R is NOT differentiable at f(x_0).
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Since f^-1[f(x)] = x, we have that, by the Chain Rule:
f'(x) * (f^-1)'[f(x)] = 1, by differentiating both sides
==> (f^-1)'[f(x)] = 1/f'(x).
When f'(x) = 0, we are dividing by zero. Since f'(x₀) = 0, we will be dividing by zero when calculating (f^-1)'[f(x₀)]. Therefore, (f^-1)'[f(x₀)] is not defined and hence not differentiable at f(x₀).
I hope this helps!
f'(x) * (f^-1)'[f(x)] = 1, by differentiating both sides
==> (f^-1)'[f(x)] = 1/f'(x).
When f'(x) = 0, we are dividing by zero. Since f'(x₀) = 0, we will be dividing by zero when calculating (f^-1)'[f(x₀)]. Therefore, (f^-1)'[f(x₀)] is not defined and hence not differentiable at f(x₀).
I hope this helps!