If the functions f and g are defined by f(x) | vedclass

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(C) Given $f(x) = 3x - 4$. Let $f(x) = y$,then $y = 3x - 4$. Solving for $x$,we get $x = \frac{y + 4}{3}$. Thus,$f^{-1}(y) = \frac{y + 4}{3}$.Given $g

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$\frac{1}{3}$

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(C) Given $f(x) = 3x - 4$. Let $f(x) = y$,then $y = 3x - 4$. Solving for $x$,we get $x = \frac{y + 4}{3}$. Thus,$f^{-1}(y) = \frac{y + 4}{3}$.Given $g(x) = 2 + 3x$. Let $g(x) = z$,then $z = 2 + 3x$. Solving for $x$,we get $x = \frac{z - 2}{3}$. Thus,$g^{-1}(z) = \frac{z - 2}{3}$.First,calculate $f^{-1}(5)$:$f^{-1}(5) = \frac{5 + 4}{3} = \frac{9}{3} = 3$.Now,calculate $g^{-1}(f^{-1}(5)) = g^{-1}(3)$:$g^{-1}(3) = \frac{3 - 2}{3} = \frac{1}{3}$.Therefore,the value is $\frac{1}{3}$.

If the functions $f$ and $g$ are defined by $f(x) = 3x - 4$ and $g(x) = 2 + 3x$ for $x \in R$,then $g^{-1}(f^{-1}(5))$ is equal to

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