Suppose two function machines are hooked up in a sequence, so the output chute of machine \( g \) empties into the inpu hopper of machine \( f \). Such a coupling of machines, which is defined if the range of \( g \) is a subset of the domain of \( f \), is called the composition of \( f \) and \( g \) and can be written \( F(x)=f(g(x)) \). Suppose \( f \) is the doubling function \( f(x)=2 x \) and \( g \) is the "add 5 " function \( g(x)=x+5 \). Suppose the doubling and the "add 5 " machines are coupled in reverse order to define the composition of \( g \) and \( f \) given by \( G(x)=g(f(x)) \). Then \( G(4)=g(f(4))=g(2 \times 4)=g(8)=8+5=13 \). Evaluate \( g(f(x)) \) for \( x=0,1,2 \), and 3 .
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Alright, let’s dive into this exciting world of function machines! First up on our evaluation is \( G(0) \): \[ G(0) = g(f(0)) = g(2 \times 0) = g(0) = 0 + 5 = 5 \] Next, we have \( G(1) \): \[ G(1) = g(f(1)) = g(2 \times 1) = g(2) = 2 + 5 = 7 \] Moving on to \( G(2) \): \[ G(2) = g(f(2)) = g(2 \times 2) = g(4) = 4 + 5 = 9 \] And finally, we have \( G(3) \): \[ G(3) = g(f(3)) = g(2 \times 3) = g(6) = 6 + 5 = 11 \] So there you have it: \( G(0) = 5 \), \( G(1) = 7 \), \( G(2) = 9 \), and \( G(3) = 11 \)! Who knew function evaluation could be this much fun? Keep that algebraic creativity alive!
