Inverse Functions — Question 1

Question 1 - Solution

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

A function is one-to-one if it passes the Horizontal Line Test, or algebraically if $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$.

Suppose: $$\frac{2x_1 + 5}{x_1 - 3} = \frac{2x_2 + 5}{x_2 - 3}$$

Cross-multiplying: $$(2x_1 + 5)(x_2 - 3) = (2x_2 + 5)(x_1 - 3)$$

Expanding both sides: $$2x_1x_2 - 6x_1 + 5x_2 - 15 = 2x_1x_2 - 6x_2 + 5x_1 - 15$$

Subtract $2x_1x_2$ and $-15$ from both sides: $$-6x_1 + 5x_2 = -6x_2 + 5x_1 \Rightarrow 5x_2 + 6x_2 = 5x_1 + 6x_1 \Rightarrow 11x_2 = 11x_1 \Rightarrow x_1 = x_2$$

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

(b) Find the inverse function:

Let $y = \dfrac{2x + 5}{x - 3}$, and solve for $x$:

$$y(x - 3) = 2x + 5 \Rightarrow yx - 3y = 2x + 5 \Rightarrow yx - 2x = 3y + 5 \Rightarrow x(y - 2) = 3y + 5 \Rightarrow x = \frac{3y + 5}{y - 2}$$

Now swap $x$ and $y$:

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

(c) Domain and Range:

For $f(x) = \dfrac{2x + 5}{x - 3}$: - Denominator cannot be 0 ⇒ $x \neq 3$ - So domain of $f$ is: $\boxed{(-\infty, 3) \cup (3, \infty)}$

Range: Find $y$ values that cannot occur. From inverse function: $f^{-1}(x) = \dfrac{3x + 5}{x - 2} \Rightarrow x \neq 2$

So range of $f$ is $\boxed{(-\infty, 2) \cup (2, \infty)}$

Domain and Range: $$\begin{aligned} \text{Domain of } f(x) &: (-\infty, 3) \cup (3, \infty) \\ \text{Range of } f(x) &: (-\infty, 2) \cup (2, \infty) \\ \text{Domain of } f^{-1}(x) &: (-\infty, 2) \cup (2, \infty) \\ \text{Range of } f^{-1}(x) &: (-\infty, 3) \cup (3, \infty) \end{aligned}$$