Topic 4: Functions [MI/2014/MYE PU1/13] The function $\mathrm{f}$ is defined by
$\mathrm{f} : x \mapsto x^2 - 6x + 14, \quad x \in \mathbb{R}.$

1. Provide a counter-example by finding two distinct inputs that yield the same output.

Let $x = 0$ and $x = 6$.
$\mathrm{f}(0) = 0^2 - 6(0) + 14 = 14$
$\mathrm{f}(6) = 6^2 - 6(6) + 14 = 14$

2. State the conclusion based on the definition of a one-one function.

Since $\mathrm{f}(0) = \mathrm{f}(6)$ but $0 \neq 6$, there are distinct values in the domain that map to the same value in the codomain. Therefore, $\mathrm{f}$ is not a one-one function.

1. Express the function $\mathrm{f}(x)$ in the completed square form to locate its turning point.

$\mathrm{f}(x) = x^2 - 6x + 14$
$= (x-3)^2 - 3^2 + 14$
$= (x-3)^2 + 5$

2. Determine the maximum value of $a$ for the restricted domain.

For a quadratic function to be one-one, its domain must be restricted to one side of the line of symmetry.
The line of symmetry is at $x = 3$.
Since the domain is $x \le a$, the function will only remain one-one if it does not cross the turning point.
Therefore, the maximum value of $a$ is $3$.

3. Form an equation to find the inverse function by making $x$ the subject.

Let $y = \mathrm{f}(x)$.
$y = (x-3)^2 + 5$
$(x-3)^2 = y - 5$
$x - 3 = \pm\sqrt{y - 5}$

4. Select the correct root based on the domain restriction and write the final expression.

Given the new domain $x \le 3$, it follows that $x - 3 \le 0$.
Thus, we reject the positive square root.
$x - 3 = -\sqrt{y - 5}$
$x = 3 - \sqrt{y - 5}$
Hence, the expression for the inverse function is:
$\mathrm{f}^{-1}(x) = 3 - \sqrt{x - 5}$