Inverse Functions — Question 3

Question 3 - Solution

(a) Find the inverse of $f(x) = \dfrac{x}{x + 1}$:

Let $y = \dfrac{x}{x + 1}$. Solve for $x$:

Multiply both sides by $x + 1$: $$y(x + 1) = x \Rightarrow yx + y = x \Rightarrow yx - x = -y \Rightarrow x(y - 1) = -y \Rightarrow x = \frac{-y}{y - 1}$$

Now switch $x$ and $y$: $$f^{-1}(x) = \frac{-x}{x - 1}$$

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

(b) Domain and Range:

For $f(x) = \dfrac{x}{x + 1}$: Domain: $x \neq -1$ , As $x \to \infty$, $f(x) \to 1$, and as $x \to -\infty$, $f(x) \to 1$ , Solve $\dfrac{x}{x + 1} = 1 \Rightarrow x = x + 1 \Rightarrow 0 = 1$ → contradiction

So $f(x) \neq 1$

Domain: $\boxed{(-\infty, -1) \cup (-1, \infty)}$ Range: $\boxed{(-\infty, 1) \cup (1, \infty)}$

Now for $f^{-1}(x) = \dfrac{-x}{x - 1}$:

Domain excludes $x = 1$, since denominator is zero Range excludes $y = -1$, since solving $\frac{-x}{x - 1} = -1 \Rightarrow x = \frac{1}{2}$, which contradicts range of original function

Domain of $f^{-1}(x) = \boxed{(-\infty, 1) \cup (1, \infty)}$ Range of $f^{-1}(x) = \boxed{(-\infty, -1) \cup (-1, \infty)}$

(c) Verify:

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

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