Suppose f(−2)=−8\displaystyle {f{{\left(-{2}\right)}}}=-{8}, and f′(−2)=−27\displaystyle {f}'{\left(-{2}\right)}=-\frac{{2}}{{7}}.

(f−1)′(−8)=\displaystyle {\left({f}^{{-{{1}}}}\right)}'{\left(-{8}\right)}=  

Write the equation of the tangent line to f−1(x)\displaystyle {{f}^{{-{{1}}}}{\left({x}\right)}} at x=−8\displaystyle {x}=-{8}.
y=\displaystyle {y}=  

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