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

1. For the piece $x \leq 2$, we have: $(x - 2)^2 + 1 > 28 \Rightarrow (x - 2)^2 > 27 \Rightarrow x - 2 > \sqrt{27} \text{ or } x - 2 < -\sqrt{27}$ This leads to two cases:

   • $x > 2 + \sqrt{27} \Rightarrow x > 2 + 3\sqrt{3}$
   • $x < 2 - \sqrt{27} \Rightarrow x < -1.2$ (not viable here since we need $x > 2$)

2. For the second piece $x > 2$: $\frac{4x - 7}{x} > 28 \Rightarrow 4x - 7 > 28x \Rightarrow 24x < 7 \Rightarrow x < \frac{7}{24}$ But this is not true for $x > 2$.

3. Therefore, the final solution for this part is: $x \in \{x: x < 2 - \sqrt{27} \text{ or } x > 2 + 3\sqrt{3}\}$

For $h$, we note that:

• It is a one-to-one function since it is defined as a quadratic function shifted up, which is a concave-up parabola. It has a minimum point and does not reuse y-values.

Conversely, for $g$:

• It is a many-to-one function due to the quadratic nature of its first part, meaning multiple values of $x$ yield the same $g(x)$. This fail to satisfy the horizontal line test, indicating it does not have an inverse.

1. To find $h^{-1}(y) = \frac{1}{2}$, we begin from the definition of $h(x)$: $h(x) = (x - 2)^2 + 1$ Set this equal to rac{1}{2}: $(x - 2)^2 + 1 = \frac{1}{2} \Rightarrow (x - 2)^2 = \frac{1}{2} - 1 = -\frac{1}{2}$

2. This equation has no real solutions.

Thus, there are no values of $x$ that satisfy the equation $h^{-1}(y) = \frac{1}{2}$.