If $\mathrm{f}\left(\mathrm{x}\right)=-4{\mathrm{e}}^{\frac{1-\mathrm{x}}{2}}+\frac{{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^2}{2}+\mathrm{x}+1\text{ and }\mathrm{g}\left(\mathrm{x}\right)={\mathrm{f}}^{-1}\left(\mathrm{x}\right)\text{ then }{\mathrm{g}}^1(-7/6)\text{ is }$

$\begin{array}{l}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{f}}^{-1}\left(\mathrm{x}\right)\Rightarrow \mathrm{f}\circ \mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}\\ differentiate w.r.to \text{'}x\text{'} on both sides\\ \Rightarrow f\text{'}\left(g\right(x\left)\right)g\text{'}\left(x\right)=1\\ \Rightarrow {\mathrm{g}}^1\left(\mathrm{x}\right)=\frac{1}{{\mathrm{f}}^1\left(\mathrm{g}\right(\mathrm{x}\left)\right)}\to \left(1\right)\\ Replacing f\left(x\right)=\frac{-7}{6}\\ -4e^{\frac{1-x}{2}}+\frac{x^3}{3}+\frac{x^2}{2}+x+1=\frac{-7}{6}\\ \Rightarrow -4e^{\frac{1-x}{2}}+\frac{x^3}{3}+\frac{x^2}{2}+x=\frac{-13}{6}\\ \Rightarrow x=1\\ \mathrm{From} \left(1\right) {\mathrm{g}}^1(-7/6)=\frac{1}{{\mathrm{f}}^1\left(\mathrm{g}\right(-7/6\left)\right)}=\frac{1}{{\mathrm{f}}^1\left(1\right)}=\frac{1}{5}\end{array}$