Suppose f(1)=0f(1) = 0f(1)=0, f′(1)=5f'(1) = 5f′(1)=5. Then (f−1)′(0)=(f^{-1})'(0) =(f−1)′(0)=A000B555CUndefinedD15\tfrac{1}{5}51check_circleExplanation(f−1)′(0)=1/f′(f−1(0))=1/f′(1)=1/5(f^{-1})'(0) = 1/f'(f^{-1}(0)) = 1/f'(1) = 1/5(f−1)′(0)=1/f′(f−1(0))=1/f′(1)=1/5.