Given the functions
[tex]\begin{gathered} f(x)=3x+9 \\ g(x)=\frac{1}{3}x-3 \end{gathered}[/tex]
To decide whether or not the functions are inverses of each other,
Solving f(x) inversely i.e to give f⁻¹(x)
Where f(x) = y
[tex]y=3x+9[/tex]
Replace y with x and x with y to give
[tex]x=3y+9[/tex]
Solve for y i.e make y the subject
[tex]\begin{gathered} x=3y+9 \\ 3y=x-9 \\ \text{Divide both sides by 3} \\ \frac{3y}{3}=\frac{x-9}{3} \\ y=\frac{x}{3}-\frac{9}{3} \\ y=\frac{1}{3}x-3 \end{gathered}[/tex]
The inverse of f(x) i.e f⁻¹(x) is
[tex]f^{-1}(x)=\frac{1}{3}x-3[/tex]
From the above deductions, it can be seen that g(x) is the inverse of f(x), i.e g(x) = f⁻¹(x)
[tex]g(x)=f^{-1}(x)=\frac{1}{3}x-3[/tex]
Thus, the functions are inverse of each other.
The answer is Yes.
