Given f(x)=6x+2\displaystyle {f{{\left({x}\right)}}}={6}{x}+{2}f(x)=6x+2 and g(x)=x+2\displaystyle {g{{\left({x}\right)}}}=\sqrt{{{x}}}+{2}g(x)=x+2, find f(g−1(x))=\displaystyle {f{{\left({{g}^{{-{{1}}}}{\left({x}\right)}}\right)}}}=f(g−1(x))= .
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