The function f is defined by f : $x \mapsto 2x^2 - 8x + 11$ for $x \in \mathbb{R}$. (i) Express f(x) in the form $a(x+b)^2 + c$, where a, b and c are constants. (ii) State the range of f. (iii) Explain why f does not have an inverse. The function g is defined by g : $x \mapsto 2x^2 - 8x + 11$ for $x \le A$, where A is a constant. (iv) State the largest value of A for which g has an inverse. (v) When A has this value, obtain an expression, in terms of x, for g^{-1}(x) and state the range of g^{-1}.

1. Factor out the coefficient of $x^2$ from the $x^2$ and $x$ terms.

$f(x) = 2(x^2 - 4x) + 11$

2. Complete the square for the expression inside the bracket.

$= 2 \left[ \left(x - \frac{4}{2}\right)^2 - \left(\frac{4}{2}\right)^2 \right] + 11$
$= 2 \left[ (x - 2)^2 - 4 \right] + 11$

3. Expand and simplify to obtain the final form.

$= 2(x - 2)^2 - 8 + 11$
$= 2(x - 2)^2 + 3$

1. Determine the range using the completed square form from part (i)

First, we evaluate the expression from part (i):
$f(x) = 2x^2 - 8x + 11 = 2(x-2)^2 + 3$

Since $(x-2)^2 \ge 0$ for all $x \in \mathbb{R}$, we have:
$2(x-2)^2 + 3 \ge 3$

Therefore, the range is $f(x) \ge 3$.

1. State that the function is not one-to-one

Since $f(x)$ is a quadratic function defined for all real $x$, it is a many-to-one function. For an inverse function to exist, the function must be one-to-one.

1. Identify the vertex from the completed square form derived in part (i).

From part (i), $g(x) = 2(x - 2)^2 + 3$.
The graph is a parabola with a minimum point (vertex) at $x = 2$, which is also its line of symmetry.

2. Determine the largest value of A for the function to be one-to-one.

For $g$ to have an inverse, it must be a one-to-one function.
For the domain $x \le A$, the graph must not cross the line of symmetry. Thus, the largest possible domain ending at $A$ without crossing the minimum point is when $A$ is exactly at the line of symmetry.
Therefore, $A = 2$.

1. Set up the equation to find the inverse, using the completed square form from part (i).

From part (i), we have $g(x) = 2(x-2)^2 + 3$.
From part (iv), the largest value is $A = 2$, so the domain is $x \le 2$.
Let $y = 2(x-2)^2 + 3$.
$\frac{y-3}{2} = (x-2)^2$

2. Take the square root of both sides and choose the correct sign based on the restricted domain.

$x - 2 = \pm\sqrt{\frac{y-3}{2}}$
Since the domain of $g$ is $x \le 2$, we have $x - 2 \le 0$.
Therefore, we reject the positive root and take the negative root:
$x - 2 = -\sqrt{\frac{y-3}{2}}$

3. Make $x$ the subject and express $g^{-1}(x)$ in terms of $x$.

$x = 2 - \sqrt{\frac{y-3}{2}}$
Hence, $g^{-1}(x) = 2 - \sqrt{\frac{x-3}{2}}$.

4. State the range of the inverse function.

The range of an inverse function $g^{-1}$ is exactly the domain of its original function $g$.
Since the domain of $g$ is $x \le 2$, the range of $g^{-1}$ is $g^{-1}(x) \le 2$.