Answer:
Step-by-step explanation:
To do this, we will use a u substitution. Namely, let 5x + 2 = u and solve for x so u is in terms of x, since we have x in our composition.
5x + 2 = u, then
5x = u - 2 and
[tex]x=\frac{u-2}{5}[/tex]
which changes f(g(x)) into f(u) {since g(x) = 5x + 2 = u}:
[tex]f(u)=\frac{2(\frac{u-2}{5})-1 }{\frac{u-2}{5} }[/tex] and of course that mess needs to be simplified:
[tex]f(u)=\frac{\frac{2u-4}{5}-1 }{\frac{u-2}{5} }[/tex] and
[tex]f(u)=\frac{\frac{2u-4}{5}-\frac{5}{5} }{\frac{u-2}{5} }[/tex] and
[tex]f(u)=\frac{\frac{2u-4-5}{5} }{\frac{u-2}{5} }[/tex] and
[tex]f(u)=\frac{\frac{2u-9}{5} }{\frac{u-2}{5} }[/tex] and bring up the lower fraction and flip it to multiply to get
[tex]f(u)=\frac{2u-9}{u-2}[/tex] and now we can change the u to an x to get that
[tex]f(x)=\frac{2x-9}{x-2}[/tex]
