Figure 4 shows a sketch of the graph of $y = g(x)$, where g(x) = \begin{cases} (x - 2)^2 + 1 & \text{for } x \leq 2 \\ \frac{4x - 7}{x > 2} \end{cases} (a) Find the value of gg(0) - Edexcel - A-Level Maths Pure - Question 7 - 2019 - Paper 2

To solve $g(x) > 28$, we need to consider both parts of the piecewise function.

1. For $x \leq 2$: $g(x) = (x - 2)^2 + 1 > 28$ $\Rightarrow (x - 2)^2 > 27$ $\Rightarrow |x - 2| > \sqrt{27}$ This gives us two inequalities: $x - 2 < -\sqrt{27}, \text{ or } x - 2 > \sqrt{27}$ Hence: $x < 2 - 3\sqrt{3}, \text{ or } x > 2 + 3\sqrt{3}$

2. For $x > 2$: $g(x) = \frac{4x - 7}{x > 2} > 28$ $\Rightarrow 4x - 7 > 28$ $\Rightarrow 4x > 35$ $\Rightarrow x > 8.75$

Thus, the combined solution is: $\{x | x < 2 - 3\sqrt{3} \text{ or } x > 8.75\}.$

The function h is one-to-one, which means that for every value of y, there is a unique value of x such that h(x) = y. This is evident since h is defined as a parabola that opens upwards but translated, hence it passes the horizontal line test.

Conversely, the function g is not one-to-one because it has a minimum point and then continues increasing; thus, a horizontal line can intersect the graph at two points, indicating that g does not have an inverse.

To solve for $h^{-1}(y) = \frac{1}{2}$, we first need to find h(x) such that:

$h(x) = \frac{1}{2}$

For $x \leq 2$,

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

This implies:

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

Since the left side cannot be negative, we find that there are no solutions in this case. Therefore, there are no x values where h(x) equals 0.5, so we return to the inverse context to deduce that the function could trace back to $x = 7.25$ using exploration of the domain and values provided.