Question

Determine whether the function is one-to-one. If it is, find its inverse function. (If an answer does not exist, enter DNE.)

Answer 1

Given the function:

`f(x)=ax+b,\text{ a}\\e0`

For each value of x, then there is only one value for f(x). Thus, the function is one-to-one.

Then, let's find the inverse function.

To find the inverse, substitute f(x) by x in x by 0:

`x=ay+b`

Now, solve for y by subtracting b from both sides:

`\begin{gathered} x-b=ay+b-b \\ x-b=ay \end{gathered}`

And divide both sides by a:

`\begin{gathered} (x-b)/(a)=(a)/(a)y \\ (x-b)/(a)=y \\ y=(x-b)/(a) \end{gathered}`

Finally, substitute y by f⁻¹(x):

`f^(-1)(x)=(x-b)/(a)`

Answer: The function is one-to-one and the inverse is:

`f^(-1)(x)=(x-b)/(a)`