One way to evaluate f(g(3))f(g(3))f, left parenthesis, g, left parenthesis, 3, right parenthesis, right parenthesis is to work from the "inside out". In other words, let's evaluate g(3)g(3)g, left parenthesis, 3, right parenthesis first and then substitute that result into fff to find our answer.
One way to evaluate f(g(3))f(g(3))f, left parenthesis, g, left parenthesis, 3, right parenthesis, right parenthesis is to work from the "inside out". In other words, let's evaluate g(3)g(3)g, left parenthesis, 3, right parenthesis first and then substitute that result into fff to find our answer.Let's evaluate g({3})g(3)g, left parenthesis, 3, right parenthesis.
One way to evaluate f(g(3))f(g(3))f, left parenthesis, g, left parenthesis, 3, right parenthesis, right parenthesis is to work from the "inside out". In other words, let's evaluate g(3)g(3)g, left parenthesis, 3, right parenthesis first and then substitute that result into fff to find our answer.Let's evaluate g({3})g(3)g, left parenthesis, 3, right parenthesis.\begin{aligned}g(x)&=x^3+2 g(3)&=({3})^3 +2 small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned}
small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned} g(x)
small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned} g(x)g(3)
small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned} g(x)g(3)
small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned} g(x)g(3)
small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned} g(x)g(3) =x
small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned} g(x)g(3) =x 3
small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned} g(x)g(3) =x 3 +2
small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned} g(x)g(3) =x 3 +2=(3)
small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned} g(x)g(3) =x 3 +2=(3) 3
small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned} g(x)g(3) =x 3 +2=(3) 3 +2 Plug in x=3.
small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned} g(x)g(3) =x 3 +2=(3) 3 +2 Plug in x=3.=29
small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned} g(x)g(3) =x 3 +2=(3) 3 +2 Plug in x=3.=29
small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned} g(x)g(3) =x 3 +2=(3) 3 +2 Plug in x=3.=29
small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned} g(x)g(3) =x 3 +2=(3) 3 +2 Plug in x=3.=29 Since g(3)=29g(3)=29g, left parenthesis, 3, right parenthesis, equals, 29, then f(g(3))=f(29)f(g(3))=f(29)f, left parenthesis, g, left parenthesis, 3, right parenthesis, right parenthesis, equals, f, left parenthesis, 29, right parenthesis.
small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned} g(x)g(3) =x 3 +2=(3) 3 +2 Plug in x=3.=29 Since g(3)=29g(3)=29g, left parenthesis, 3, right parenthesis, equals, 29, then f(g(3))=f(29)f(g(3))=f(29)f, left parenthesis, g, left parenthesis, 3, right parenthesis, right parenthesis, equals, f, left parenthesis, 29, right parenthesis.Now let's evaluate f({29})f(29)f, left parenthesis, 29, right parenthesis.
small{\gray{\text{Plug in }x={3.}}} &={29}\end{aligned} g(x)g(3) =x 3 +2=(3) 3 +2 Plug in x=3.=29 Since g(3)=29g(3)=29g, left parenthesis, 3, right parenthesis, equals, 29, then f(g(3))=f(29)f(g(3))=f(29)f, left parenthesis, g, left parenthesis, 3, right parenthesis, right parenthesis, equals, f, left parenthesis, 29, right parenthesis.Now let's evaluate f({29})f(29)f, left parenthesis, 29, right parenthesis.\begin{aligned}f(x)&=3x-1 f( {{29}})&=3({29}) - 1 small{\gray{\text{Plug in }x= {29.}}} &={86} \end{aligned}
small{\gray{\text{Plug in }x= {29.}}} &={86} \end{aligned} f(x)
small{\gray{\text{Plug in }x= {29.}}} &={86} \end{aligned} f(x)f(29)
small{\gray{\text{Plug in }x= {29.}}} &={86} \end{aligned} f(x)f(29)
small{\gray{\text{Plug in }x= {29.}}} &={86} \end{aligned} f(x)f(29)
small{\gray{\text{Plug in }x= {29.}}} &={86} \end{aligned} f(x)f(29) =3x−1
small{\gray{\text{Plug in }x= {29.}}} &={86} \end{aligned} f(x)f(29) =3x−1=3(29)−1 Plug in x=29.
small{\gray{\text{Plug in }x= {29.}}} &={86} \end{aligned} f(x)f(29) =3x−1=3(29)−1 Plug in x=29.=86
small{\gray{\text{Plug in }x= {29.}}} &={86} \end{aligned} f(x)f(29) =3x−1=3(29)−1 Plug in x=29.=86
small{\gray{\text{Plug in }x= {29.}}} &={86} \end{aligned} f(x)f(29) =3x−1=3(29)−1 Plug in x=29.=86
small{\gray{\text{Plug in }x= {29.}}} &={86} \end{aligned} f(x)f(29) =3x−1=3(29)−1 Plug in x=29.=86 It follows that f(g({3}))=f( {29})={86}f(g(3))=f(29)=86f, left parenthesis, g, left parenthesis, 3, right parenthesis, right parenthesis, equals, f, left parenthesis, 29, right parenthesis, equals, 86.
