Inverse Functions — Question 5

Question 5 - Solution

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

Suppose $f(x_1) = f(x_2)$. Then: $$\frac{3x_1 - 4}{5x_1 + 2} = \frac{3x_2 - 4}{5x_2 + 2}$$

Cross-multiplying: $$(3x_1 - 4)(5x_2 + 2) = (3x_2 - 4)(5x_1 + 2)$$

Expanding both sides: $$15x_1x_2 + 6x_1 - 20x_2 - 8 = 15x_1x_2 + 6x_2 - 20x_1 - 8$$

Cancel $15x_1x_2$ and $-8$: $$6x_1 - 20x_2 = 6x_2 - 20x_1 \Rightarrow 6x_1 + 20x_1 = 6x_2 + 20x_2 \Rightarrow 26x_1 = 26x_2 \Rightarrow x_1 = x_2$$

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

(b) Find the inverse function:

Let $y = \frac{3x - 4}{5x + 2}$

Multiply both sides: $$y(5x + 2) = 3x - 4 \Rightarrow 5xy + 2y = 3x - 4 \Rightarrow 5xy - 3x = -2y - 4 \Rightarrow x(5y - 3) = -2y - 4 \Rightarrow x = \frac{-2y - 4}{5y - 3}$$

Now swap $x$ and $y$: $$f^{-1}(x) = \frac{-2x - 4}{5x - 3}$$

(c) Domain and Range:

The only restriction on the domain of $f$ is $5x + 2 \neq 0 \Rightarrow x \neq -\frac{2}{5}$

The inverse is undefined when the denominator is 0: $$5x - 3 \neq 0 \Rightarrow x \neq \frac{3}{5}$$

So:

$$\text{Domain of } f(x): \boxed{(-\infty, -\tfrac{2}{5}) \cup (-\tfrac{2}{5}, \infty)}$$ $$\text{Range of } f(x): \boxed{(-\infty, \tfrac{3}{5}) \cup (\tfrac{3}{5}, \infty)}$$

$$\text{Domain of } f^{-1}(x): \boxed{(-\infty, \tfrac{3}{5}) \cup (\tfrac{3}{5}, \infty)}$$ $$\text{Range of } f^{-1}(x): \boxed{(-\infty, -\tfrac{2}{5}) \cup (-\tfrac{2}{5}, \infty)}$$