Question

Inverse Function In Exercise,analytically show that the functions are inverse functions.Then use the graphing utility to show this graphically.
f(x) = e^2x-1
g(x) = 1/2 + In(x)1/2

Answer 1

Answer:

$f^{-1}(x)=g(x)=\frac{\text{ln}(x)}{2}+\frac{1}{2}$  

Step-by-step explanation:

Please find the attachment.

We have been given two functions as $(x)=e^{2x-1}$ and $g(x)=\frac{1}{2}+\frac{\text{ln}(x)}{2}$. We are asked to show that both functions are inverse of each other algebraically and graphically.

Let us find inverse function of $f(x)=e^{2x-1}$ as:

$y=e^{2x-1}$

Interchange x and y values:  

$x=e^{2y-1}$

Take natural log of both sides:

$\text{ln}(x)=\text{ln}(e^{2y-1})$  

$\text{ln}(x)=(2y-1)*\text{ln}(e)$  

$\text{ln}(x)=(2y-1)*1$  

$\text{ln}(x)=2y-1$  

$\text{ln}(x)+1=2y-1+1$

$\text{ln}(x)+1=2y$  

$\frac{\text{ln}(x)+1}{2}=\frac{2y}{2}$  

$\frac{\text{ln}(x)}{2}+\frac{1}{2}=y$  

$f^{-1}(x)=\frac{\text{ln}(x)}{2}+\frac{1}{2}$  

Therefore, we can see that function $g(x)=\frac{1}{2}+\frac{\text{ln}(x)}{2}$ is inverse of function $f(x)=e^{2x-1}$.

We can see that both functions are symmetric about line $y=x$, therefore, both functions are inverse of each other.