Inverse Functions — Question 8

Question 8 - Solution

(a) Show that $f$ is one-to-one:

Let $f(x_1) = f(x_2)$:

$$\frac{x_1 + 3}{x_1 - 1} = \frac{x_2 + 3}{x_2 - 1} \Rightarrow (x_1 + 3)(x_2 - 1) = (x_2 + 3)(x_1 - 1)$$

Expanding both sides:

$$x_1x_2 - x_1 + 3x_2 - 3 = x_1x_2 - x_2 + 3x_1 - 3$$

Subtract $x_1x_2$ and $-3$ from both sides:

$$- x_1 + 3x_2 = - x_2 + 3x_1 \Rightarrow 3x_2 + x_2 = 3x_1 + x_1 \Rightarrow 4x_2 = 4x_1 \Rightarrow x_1 = x_2$$

Therefore, $f$ is one-to-one.

(b) Find the inverse:

Let $y = \dfrac{x + 3}{x - 1}$

Multiply both sides:

$$y(x - 1) = x + 3 \Rightarrow yx - y = x + 3 \Rightarrow yx - x = y + 3 \Rightarrow x(y - 1) = y + 3 \Rightarrow x = \frac{y + 3}{y - 1}$$

Switch $x$ and $y$:

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

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

Interestingly, $f^{-1}(x) = f(x)$. The function is its own inverse.

(c) Domain and Range:

From $f(x) = \dfrac{x + 3}{x - 1}$, we see that: - Domain: $x \neq 1 \Rightarrow \boxed{(-\infty, 1) \cup (1, \infty)}$ - As $x \to \infty$, $f(x) \to 1$, and $f(x) \neq 1$ - So range: $\boxed{(-\infty, 1) \cup (1, \infty)}$

The same applies to $f^{-1}(x)$, so: - Domain of $f^{-1}(x): (-\infty, 1) \cup (1, \infty)$ - Range of $f^{-1}(x): (-\infty, 1) \cup (1, \infty)$

(d) Verify:

$$f(f^{-1}(x)) = f\left( \frac{x + 3}{x - 1} \right) = \frac{\left( \frac{x + 3}{x - 1} \right) + 3}{\left( \frac{x + 3}{x - 1} \right) - 1} = \frac{\frac{x + 3 + 3(x - 1)}{x - 1}}{\frac{x + 3 - (x - 1)}{x - 1}} = \frac{\frac{x + 3 + 3x - 3}{x - 1}}{\frac{x + 3 - x + 1}{x - 1}} = \frac{\frac{4x}{x - 1}}{\frac{4}{x - 1}} = x$$

Verified: $f(f^{-1}(x)) = x$