Inverse Functions

For $f : (-\infty, 1] \to R$, $f(x) = x^2 - 2x$, determine the equation and the domain for the inverse function $f^{-1}$.

1. 1

   Write $f$ in turning-point form and swap $x$ and $y$.

   $$x = (y-1)^2 - 1$$
2. 2

   Solve for $y$: $(y-1)^2 = x+1$, so $y = 1 \pm \sqrt{x+1}$.

   $$y = 1 \pm \sqrt{x+1}$$
3. 3

   Since the original domain is $(-\infty,1]$, the inverse range is $(-\infty,1]$, so take the negative root. The domain of $f^{-1}$ is the range of $f$. The report notes the most common error was choosing the positive arm $1+\sqrt{x+1}$.

   $$f^{-1}(x) = 1 - \sqrt{x+1},\quad \text{domain } [-1,\infty)$$

VCAA 2023 Mathematical Methods Exam 1, Q7c

Let $g(x) = e^{2x} - 8e^{x} + 7$, where $x \in R$. The function $g(x)$ has exactly one stationary point, a local minimum. Find the largest value of $a$ such that when $g$ is restricted to the domain $(-\infty, a]$ it has an inverse function.

1. 1

   Differentiate and set equal to zero to locate the local minimum.

   $$g'(x) = 2e^{2x} - 8e^{x} = 2e^{x}(e^{x} - 4) = 0$$
2. 2

   Solve for $x$ (since $e^x>0$).

   $$e^{x} = 4 \implies x = \log_e(4)$$
3. 3

   The largest domain on which $g$ is one-to-one ends at the turning point. The report warns against letting $u=e^x$, finding $u=4$, then wrongly taking $a=4$ instead of $a=\log_e 4$.

   $$a = \log_e(4)$$

VCAA 2025 Mathematical Methods Exam 1, Q5b

Find the rule for the inverse of $f(x) = 2x - 6$, and state its domain and range. The domain of $f$ is $R$.

1. 1

   Write $y$ for $f(x)$, then swap $x$ and $y$. Swapping is the whole trick.

   $$y = 2x - 6 \quad\Rightarrow\quad x = 2y - 6$$
2. 2

   Make $y$ the subject again.

   $$x + 6 = 2y \quad\Rightarrow\quad y = \frac{x + 6}{2}$$
3. 3

   The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$. Here $f$ has range $R$ and domain $R$.

   $$f^{-1}(x) = \frac{x + 6}{2}, \quad \text{domain } R, \quad \text{range } R$$

The function $g(x) = x^2 - 4$ has domain $[0, \infty)$. Find $g^{-1}$, and state its domain and range.

1. 1

   On $[0, \infty)$ the parabola is one to one, so an inverse function exists. Find the range of $g$ first, because it becomes the domain of $g^{-1}$.

   $$\text{range of } g = [-4, \infty)$$
2. 2

   Write $y = x^2 - 4$, swap, then make $y$ the subject. Keep only the branch that matches the original domain.

   $$x = y^2 - 4 \quad\Rightarrow\quad y = \sqrt{x + 4}$$
3. 3

   Take the positive root because the range of $g^{-1}$ must equal the domain of $g$, which was $[0, \infty)$.

   $$g^{-1}(x) = \sqrt{x + 4}$$
4. 4

   State both sets. Domain of $g^{-1}$ is the range of $g$; range of $g^{-1}$ is the domain of $g$.

   $$\text{domain } [-4, \infty), \quad \text{range } [0, \infty)$$

The graph of $f(x) = \sqrt{x}$ meets the graph of $f^{-1}$. Find the coordinates of every point of intersection.

1. 1

   The graphs of $f$ and $f^{-1}$ are mirror images in the line $y = x$. So any intersection sits on that mirror line. Solve $f(x) = x$ instead of $f(x) = f^{-1}(x)$.

   $$\sqrt{x} = x$$
2. 2

   Square both sides and solve.

   $$x = x^2 \quad\Rightarrow\quad x^2 - x = 0 \quad\Rightarrow\quad x(x - 1) = 0$$
3. 3

   So $x = 0$ or $x = 1$. Both lie in the domain $x \ge 0$. Give full coordinates, not just the $x$ values.

   $$x = 0 \ \text{ or } \ x = 1$$