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

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

Explanation:

Please find the attachment.

We have been given two functions as
`(x)=e^(2x-1)` and
`g(x)=(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}=(2y)/(2)`

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

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

Therefore, we can see that function
`g(x)=(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.