Given the function:
[tex]f(x)=ax+b,\text{ a}\ne0[/tex]
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:
[tex]x=ay+b[/tex]
Now, solve for y by subtracting b from both sides:
[tex]\begin{gathered} x-b=ay+b-b \\ x-b=ay \end{gathered}[/tex]
And divide both sides by a:
[tex]\begin{gathered} \frac{x-b}{a}=\frac{a}{a}y \\ \frac{x-b}{a}=y \\ y=\frac{x-b}{a} \end{gathered}[/tex]
Finally, substitute y by f⁻¹(x):
[tex]f^{-1}(x)=\frac{x-b}{a}[/tex]
Answer: The function is one-to-one and the inverse is:
[tex]f^{-1}(x)=\frac{x-b}{a}[/tex]
