If the function f(x) = -4e^{left(frac{1-x}{2} | vedclass

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(A) Given $f(x) = -4e^{\left(\frac{1-x}{2}\right)} + 1 + x + \frac{x^2}{2} + \frac{x^3}{3}$.Since $g(x) = f^{-1}(x)$,we have $g(f(x)) = x$.Differentia

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$\frac{1}{5}$

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(A) Given $f(x) = -4e^{\left(\frac{1-x}{2}\right)} + 1 + x + \frac{x^2}{2} + \frac{x^3}{3}$.Since $g(x) = f^{-1}(x)$,we have $g(f(x)) = x$.Differentiating both sides with respect to $x$,we get $g'(f(x)) \cdot f'(x) = 1$,which implies $g'(f(x)) = \frac{1}{f'(x)}$.First,we find $x$ such that $f(x) = -\frac{7}{6}$. By inspection,if $x = 1$,then $f(1) = -4e^0 + 1 + 1 + \frac{1}{2} + \frac{1}{3} = -4 + 2 + \frac{3+2}{6} = -2 + \frac{5}{6} = -\frac{7}{6}$.Now,calculate $f'(x) = -4e^{\left(\frac{1-x}{2}\right)} \cdot \left(-\frac{1}{2}\right) + 1 + x + x^2 = 2e^{\left(\frac{1-x}{2}\right)} + 1 + x + x^2$.At $x = 1$,$f'(1) = 2e^0 + 1 + 1 + 1 = 2 + 3 = 5$.Therefore,$g'(-\frac{7}{6}) = g'(f(1)) = \frac{1}{f'(1)} = \frac{1}{5}$.

If the function $f(x) = -4e^{\left(\frac{1-x}{2}\right)} + 1 + x + \frac{x^2}{2} + \frac{x^3}{3}$ and $g(x) = f^{-1}(x)$,then the value of $g'(-\frac{7}{6})$ equals

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