Answer
a) Find the inverse function
$f^{-1}(x) = \frac{x+1}{3}$
b) Verify that the inverse equation is correct
$f(f^{-1}(x)) = x$
$f(\frac{x+1}{3}) = 3(\frac{x+1}{3}) -1$
$=\frac{3(x+1)}{3}-1$
$= x + 1 - 1$
$= x$
$f^{-1}(f(x)) = x$
$f^{-1}(3x-1) = \frac{(3x-1) +1}{3}$
$= \frac{3x -1 +1}{3}$
$= \frac{3x}{3}$
$= x$
Therefore, the inverse equation is correct.
Work Step by Step
a) Find the inverse equation
Let $f(x) = y$
$y = 3x - 1$
$x = 3y - 1$
$x + 1 = 3y$
$y = \frac{x+1}{3}$
$f^{-1}(x) = \frac{x+1}{3}$
b) Verify that the inverse equation is correct
$f(f^{-1}(x)) = x$
$f(\frac{x+1}{3}) = 3(\frac{x+1}{3}) -1$
$=\frac{3(x+1)}{3}-1$
$= x + 1 - 1$
$= x$
$f^{-1}(f(x)) = x$
$f^{-1}(3x-1) = \frac{(3x-1) +1}{3}$
$= \frac{3x -1 +1}{3}$
$= \frac{3x}{3}$
$= x$
Therefore, the inverse equation is correct.
