If function f(x) = - 4{e^{left( {frac{{1 - x}}{2}} right)}} + 1 + x + frac{{{x^2}}}{2} +

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5. Continuity and Differentiation

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

A

$\frac{1}{5}$

B

$- \frac{1}{5}$

C

$\frac{6}{7}$

D

$ -\frac{6}{7}$

Solution

$\because g(x)=f^{1}(x)$

${g f(x)=x}$

${g^{1} f(x) f^{\prime}(x)=1}$

$(\because f(1)=-7 / 6)$

$\therefore \mathrm{g}^{1}(\mathrm{f}(1))=\frac{1}{\mathrm{f}^{2}(1)}$

$g^{1}(-7 / 6)=\frac{1}{5}$

Standard 12

Mathematics

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