Figure 4 shows a sketch of the graph of $y = g(x)$, where $g(x) = \begin{cases} (x - 2)^2 + 1 & x \leq 2 \\ 4x - 7 & x > 2 \end{cases}$ a) Find the value of $gg(0)$ - Edexcel - A-Level Maths Pure - Question 8 - 2019 - Paper 2

To solve for x where g(x) > 28, we consider both cases of g(x).

1. For x ≤ 2:

   $(x - 2)^2 + 1 > 28$ leads to:

   $(x - 2)^2 > 27$.

   Taking the square root gives:

   $|x - 2| > \sqrt{27}$, which simplifies to:

   $x < 2 - 3\sqrt{3}$ or $x > 2 + 3\sqrt{3}$.

2. For x > 2:

   $4x - 7 > 28$ leads to:

   $4x > 35 \implies x > \frac{35}{4}$.

Combining both results, the valid intervals are:

$x < 2 - 3\sqrt{3}$ or $x > \frac{35}{4}$.

The function $h$ has an inverse because it is a one-to-one function; it passes the horizontal line test. Each input corresponds to a unique output.

Conversely, the function $g$ does not have an inverse as it is not one-to-one; for certain values of x, multiple inputs can yield the same output. This is especially evident in the case of the first expression where multiple inputs (e.g., x = 1 and x = 3) yield g(x) = 5.

To solve for x in the equation $h^{-1}(y) = \frac{1}{2}$, we first express $h(x)$:

$h(x) = (x - 2)^2 + 1$.

Setting $h(x) = \frac{1}{2}$ gives us:

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

Since $(x - 2)^2$ cannot be negative, no solutions exist. Thus:

The equation has no solution.