Answer:
[tex]f^{-1}(x)=g(x)=\frac{\text{ln}(x)}{2}+\frac{1}{2}[/tex]
Step-by-step explanation:
Please find the attachment.
We have been given two functions as [tex](x)=e^{2x-1}[/tex] and [tex]g(x)=\frac{1}{2}+\frac{\text{ln}(x)}{2}[/tex]. We are asked to show that both functions are inverse of each other algebraically and graphically.
Let us find inverse function of [tex]f(x)=e^{2x-1}[/tex] as:
[tex]y=e^{2x-1}[/tex]
Interchange x and y values:
[tex]x=e^{2y-1}[/tex]
Take natural log of both sides:
[tex]\text{ln}(x)=\text{ln}(e^{2y-1})[/tex]
[tex]\text{ln}(x)=(2y-1)*\text{ln}(e)[/tex]
[tex]\text{ln}(x)=(2y-1)*1[/tex]
[tex]\text{ln}(x)=2y-1[/tex]
[tex]\text{ln}(x)+1=2y-1+1[/tex]
[tex]\text{ln}(x)+1=2y[/tex]
[tex]\frac{\text{ln}(x)+1}{2}=\frac{2y}{2}[/tex]
[tex]\frac{\text{ln}(x)}{2}+\frac{1}{2}=y[/tex]
[tex]f^{-1}(x)=\frac{\text{ln}(x)}{2}+\frac{1}{2}[/tex]
Therefore, we can see that function [tex]g(x)=\frac{1}{2}+\frac{\text{ln}(x)}{2}[/tex] is inverse of function [tex]f(x)=e^{2x-1}[/tex].
We can see that both functions are symmetric about line [tex]y=x[/tex], therefore, both functions are inverse of each other.
