Functions: Question 3

Worked solution

Part (a): Finding $f^{-1}(x)$

Write $y = f(x)$:

$y = \frac{3x+2}{x-1}$

Swap $x$ and $y$. This is the key step that turns $f$ into $f^{-1}$:

$x = \frac{3y+2}{y-1}$

Multiply both sides by $(y-1)$:

$x(y-1) = 3y+2$

Expand the left-hand side:

$xy - x = 3y + 2$

Collect all terms containing $y$ on one side, and everything else on the other:

$xy - 3y = x + 2$

Factor $y$ out on the left:

$y(x-3) = x+2$

Divide by $(x-3)$:

$y = \frac{x+2}{x-3}$

So $f^{-1}(x) = \frac{x+2}{x-3}.$

Part (b): Domain of $f^{-1}$

The domain of $f^{-1}$ is always equal to the range of $f$.

From the working above, $f^{-1}(x)$ is undefined only when $x=3$ (division by zero) (and this is exactly the value that $f$ itself can never output. Checking directly: setting $f(x)=3$ gives $3(x-1)=3x+2$, i.e. $3x-3=3x+2$, i.e. $-3=2$, which is never true) so $y=3$ is indeed never attained by $f$, confirming it is the correct value to exclude.

So the domain of $f^{-1}$ is: $x \in \mathbb{R}, \quad x \ne 3$

Part (c): Verifying with a numerical check

First find $f(2)$:

$f(2) = \frac{3(2)+2}{2-1} = \frac{8}{1} = 8$

Now apply $f^{-1}$ to this result:

$f^{-1}(8) = \frac{8+2}{8-3} = \frac{10}{5} = 2$

Since $f^{-1}(f(2)) = f^{-1}(8) = 2$, this confirms $f^{-1}f(2) = 2$, as required. Applying $f^{-1}$ after $f$ returns the original input, exactly as an inverse function should.

Final answers

• (a) $f^{-1}(x) = \boxed{\dfrac{x+2}{x-3}}$
• (b) Domain of $f^{-1}$: $x \in \mathbb{R}, x \ne 3$
• (c) $f(2)=8$, $f^{-1}(8)=2$, so $f^{-1}f(2)=2$ ✓