Given two functions, they can be composed in two ways:
[tex](f\circ g)(x)=f(g(x))[/tex]
It reads "f compound g" or simply "we are going to fill f with g".
[tex](g\circ f)(x)=g(f(x))[/tex]
It reads "g compound f" or simply "we are going to fill g with f"
So, in this case, you have
[tex]\begin{gathered} f(x)=x^2-8x \\ g(x)=x+1 \end{gathered}[/tex]
Point a.
[tex]\begin{gathered} (f\circ g)(x)=f(g(x)) \\ (f\circ g)(x)=f(x+1) \\ (f\circ g)(x)=(x+1)^2-8(x+1) \end{gathered}[/tex]
Now, evaluating at -3
[tex]\begin{gathered} (f\circ g)(-3)=(-3+1)^2-8(-3+1) \\ (f\circ g)(-3)=(-2)^2-8(-2) \\ (f\circ g)(-3)=4+16 \\ (f\circ g)(-3)=20 \end{gathered}[/tex]
Point b.
[tex]\begin{gathered} (g\circ f)(x)=g(f(x)) \\ (g\circ f)(x)=g(x^2-8x) \\ (g\circ f)(x)=(x^2-8x)+1 \\ (g\circ f)(x)=x^2-8x+1 \end{gathered}[/tex]
Now, evaluating at -3
[tex]\begin{gathered} (g\circ f)(-3)=(-3)^2-8(-3)+1 \\ (g\circ f)(-3)=9+24+1 \\ (g\circ f)(-3)=34 \end{gathered}[/tex]
Therefore,
[tex]\begin{gathered} (f\circ g)(-3)=20 \\ \text{ and} \\ (g\circ f)(-3)=34 \end{gathered}[/tex]
