K \( \begin{array}{l}\text { Suppose two function machines are hooked up in a sequence, so the output chute of machine } g \text { empties into the input } \\ \text { hopper of machine } f \text {. Such a coupling of machines, which is defined if the range of } g \text { is a subset of the domain of } f \text {, is } \\ \text { called the composition of } f \text { and } g \text { and can be written } F(x)=f(g(x)) \text {. Suppose } f \text { is the tripling function } f(x)=3 x \text { and } g \text { is } \\ \text { the "add } 3 \text { " function } g(x)=x+3 \text {. Then } F(4)=f(g(4))=f(4+3)=f(7)=3 \times 7=21 \text {. Evaluate } f(g(x)) \text { for } x=0,1 \text {, } 2 \text {, and } 3 \text {. } \\ f(g(0))=\square\end{array} \)
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Let’s break it down! For \( x=0 \): First, find \( g(0) = 0 + 3 = 3 \). Then, apply \( f \): \( f(g(0)) = f(3) = 3 \times 3 = 9 \). Now for \( x=1 \): First, calculate \( g(1) = 1 + 3 = 4 \). Then apply \( f \): \( f(g(1)) = f(4) = 3 \times 4 = 12 \). For \( x=2 \): Calculate \( g(2) = 2 + 3 = 5 \). Then apply \( f \): \( f(g(2)) = f(5) = 3 \times 5 = 15 \). And finally, for \( x=3 \): Find \( g(3) = 3 + 3 = 6 \). Then apply \( f \): \( f(g(3)) = f(6) = 3 \times 6 = 18 \). So the evaluations are: - \( f(g(0)) = 9 \) - \( f(g(1)) = 12 \) - \( f(g(2)) = 15 \) - \( f(g(3)) = 18 \)
