Inverse Functions — Question 7

Question 7 - Solution

(a) Show $f(x) = \dfrac{6}{x + 2} - 1$ is one-to-one:

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

$$\frac{6}{x_1 + 2} - 1 = \frac{6}{x_2 + 2} - 1 \Rightarrow \frac{6}{x_1 + 2} = \frac{6}{x_2 + 2} \Rightarrow x_1 + 2 = x_2 + 2 \Rightarrow x_1 = x_2$$

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

(b) Find the inverse function:

Let $y = \dfrac{6}{x + 2} - 1$

Solve for $x$: $$y + 1 = \frac{6}{x + 2} \Rightarrow x + 2 = \frac{6}{y + 1} \Rightarrow x = \frac{6}{y + 1} - 2$$

Now switch $x$ and $y$:

$$f^{-1}(x) = \frac{6}{x + 1} - 2$$

Answer: $\boxed{f^{-1}(x) = \dfrac{6}{x + 1} - 2}$

(c) Domain and Range:

Original function $f(x) = \dfrac{6}{x + 2} - 1$

- Undefined when $x + 2 = 0 \Rightarrow x = -2$ - Domain of $f$: $\boxed{(-\infty, -2) \cup (-2, \infty)}$

Asymptote at $x = -2$, and horizontal asymptote $y = -1$ - Range of $f$: $\boxed{(-\infty, -1) \cup (-1, \infty)}$

Inverse: - Domain of $f^{-1}$: same as range of $f$: $\boxed{(-\infty, -1) \cup (-1, \infty)}$ - Range of $f^{-1}$: same as domain of $f$: $\boxed{(-\infty, -2) \cup (-2, \infty)}$