MathematicsMedium63×since 2002Q3655Let f:(−1,1)→Rf:\left( { - 1,1} \right) \to Rf:(−1,1)→R be a differentiable function with f(0)=−1f\left( 0 \right) = - 1f(0)=−1 and f′(0)=1f'\left( 0 \right) = 1f′(0)=1. Let g(x)=[f(2f(x)+2)]2g\left( x \right) = {\left[ {f\left( {2f\left( x \right) + 2} \right)} \right]^2}g(x)=[f(2f(x)+2)]2. Then g′(0)=g'\left( 0 \right) =g′(0)=A−4-4−4B000C−2-2−2D444Check answerSkip