Answer:
f^(-1)(x) = 1/5x
Step-by-step explanation:
[tex] \large{f(x) = mx + b \longrightarrow f^{ - 1} (x) = \frac{x - b}{m} }[/tex]
Inverse Function is when we swap the x-term and y-term. Here is an explanation to the inverse of linear function above.
[tex] \large{f(x) = mx + b}[/tex]
Here we define f(x) = y. It is better to use y instead of f(x)
[tex] \large{y = mx + b}[/tex]
Swap x-term and y-term.
[tex] \large{x = my + b}[/tex]
Arrange/Simplify in the y-isolated equation.
[tex] \large{x - b = my} \\ \large {\frac{x - b}{ m} = y} \\ \large{ y = \frac{x - b}{m} }[/tex]
Since the function is in inverse form. It's recommended to use f^(-1) to point out that the function is an inverse of the original function.
[tex] \large \boxed{ {f}^{ - 1} (x) = \frac{x - b}{m} }[/tex]
Hence, y = (x-b)/m is an inverse of y = mx+b. A one-to-one function can have an inverse form. Back to the question!
We are given the linear function:
[tex] \large{f(x) = 5x}[/tex]
We will be using substitution method by substituting the original function in inverse form.
Since we know that the slope is 5 and doesn't have y-intercept which is b-value. Hence
[tex] \large{ {f}^{ - 1} (x) = \frac{x - 0}{5} } \\ \large{ {f}^{ - 1} (x) = \frac{x }{5} } \\ \large{ {f}^{ - 1} (x) = \frac{1}{5} x}[/tex]
Or we swap x-term and y-term.
[tex] \large{f(x) = 5x} \\ \large{y = 5x} \\ \large{x = 5y} \\ \large{ \frac{x}{5} = y} \\ \large{y = \frac{x}{5} = \frac{1}{5} x} \\ \large{ {f}^{ - 1} (x) = \frac{1}{5} x}[/tex]
