Let R denote the set of all real numbers. Let | vedclass

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(B) Let $h(x) = f(g^{-1}(x))$. We need to find $h'(2) = f'(g^{-1}(2)) \cdot (g^{-1})'(2)$.First,find $g^{-1}(2)$. Since $g(0) = \frac{4}{1 + e^0} = \f

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$0.25$

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(B) Let $h(x) = f(g^{-1}(x))$. We need to find $h'(2) = f'(g^{-1}(2)) \cdot (g^{-1})'(2)$.First,find $g^{-1}(2)$. Since $g(0) = \frac{4}{1 + e^0} = \frac{4}{2} = 2$,we have $g^{-1}(2) = 0$.Next,find $f'(x) = \frac{2x + 2}{x^2 + 2x + 4}$. Thus,$f'(0) = \frac{2}{4} = 0.5$.Now,find $(g^{-1})'(2)$. We know $(g^{-1})'(g(x)) = \frac{1}{g'(x)}$.$g'(x) = \frac{4 \cdot (-1) \cdot e^{-2x} \cdot (-2)}{(1 + e^{-2x})^2} = \frac{8e^{-2x}}{(1 + e^{-2x})^2}$.At $x = 0$,$g'(0) = \frac{8(1)}{(1 + 1)^2} = \frac{8}{4} = 2$.Therefore,$(g^{-1})'(2) = \frac{1}{g'(0)} = \frac{1}{2} = 0.5$.Finally,$h'(2) = f'(0) \cdot (g^{-1})'(2) = 0.5 \cdot 0.5 = 0.25$.

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