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

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(C) Given functions are $f(x) = 3x - 4$ and $g(x) = 3x + 2$. To find $f^{-1}(y)$,let $f(x) = y$. $3x - 4 = y \implies 3x = y + 4 \implies x = \frac{y

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$\frac{1}{3}$

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(C) Given functions are $f(x) = 3x - 4$ and $g(x) = 3x + 2$. To find $f^{-1}(y)$,let $f(x) = y$. $3x - 4 = y \implies 3x = y + 4 \implies x = \frac{y + 4}{3}$. Thus,$f^{-1}(y) = \frac{y + 4}{3}$. Now,$f^{-1}(5) = \frac{5 + 4}{3} = \frac{9}{3} = 3$. To find $g^{-1}(z)$,let $g(x) = z$. $3x + 2 = z \implies 3x = z - 2 \implies x = \frac{z - 2}{3}$. Thus,$g^{-1}(z) = \frac{z - 2}{3}$. Finally,$g^{-1}(f^{-1}(5)) = g^{-1}(3) = \frac{3 - 2}{3} = \frac{1}{3}$.

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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