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(B) Given,$f(x) = 5x - 3$ and $g(x) = x^2 + 3$. To find $f^{-1}(x)$,let $y = f(x) = 5x - 3$. Then $y + 3 = 5x$,which implies $x = \frac{y + 3}{5}$. Th

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$\frac{111}{25}$

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(B) Given,$f(x) = 5x - 3$ and $g(x) = x^2 + 3$. To find $f^{-1}(x)$,let $y = f(x) = 5x - 3$. Then $y + 3 = 5x$,which implies $x = \frac{y + 3}{5}$. Thus,$f^{-1}(y) = \frac{y + 3}{5}$,so $f^{-1}(x) = \frac{x + 3}{5}$. Now,we need to calculate $g \circ f^{-1}(3) = g(f^{-1}(3))$. First,find $f^{-1}(3) = \frac{3 + 3}{5} = \frac{6}{5}$. Then,$g(\frac{6}{5}) = (\frac{6}{5})^2 + 3 = \frac{36}{25} + 3$. Calculating the sum: $\frac{36 + 75}{25} = \frac{111}{25}$.

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x) = 5x - 3$ and $g(x) = x^2 + 3$,then $g \circ f^{-1}(3)$ is equal to

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