Let's begin by listing out the given information:
[tex]\begin{gathered} f\mleft(x\mright)=3^x+3\Rightarrow y=3^x+3 \\ Switch\text{ }the\text{ }variables\text{ x \& y in the equation, we have:} \\ x=3^y+3\Rightarrow x-3=3^y \\ x-3=3^y\Rightarrow3^y=x-3 \\ \text{Take the }ln\text{ }o\text{f both }sides\text{, we have:} \\ y=\frac{\ln{\left(x - 3 \right)}}{\ln{\left(3 \right)}} \\ y=f^{-1}\mleft(x\mright) \\ f^{-1}\mleft(x\mright)=\frac{\ln{(x-3)}}{\ln{(3)}} \\ \end{gathered}[/tex]
2.
The domain of a function is the set of input or argument values for which the function is real and defined. This is given by:
[tex]\begin{gathered} x-3;x>3 \\ \therefore x>3 \end{gathered}[/tex]
3.
The range of a function is the set of output values for which the function is defined. This is given by:
[tex]\begin{gathered} 3^x+3\colon-\infty\:
4. The range of f(x) is the domain of f^-1(x) since they are inverse of one another.
[tex]\begin{gathered} f\mleft(x\mright)=3^x+3 \\ x=0 \\ f(0)=3^0+3=1+3=4 \\ x=1 \\ f(1)=3^1+3=3+3=6 \\ x=2 \\ f(2)=3^2+3=9+3=12 \\ \end{gathered}[/tex]
Therefore, the domain of f^-1(x) increases
5.
The asymptote of a curve is a line such that the distance between the curve and the line approaches zero
[tex]\begin{gathered} \: \frac{\ln\left(x-3\right)}{\ln\left(3\right)}\colon\quad \mathrm{Vertical}\colon\: x=3 \\ \therefore x=3 \end{gathered}[/tex]
