Figure 1 shows a sketch of part of the graph of $y = g(x)$, where g(x) = 3 + \sqrt{x + 2}, \quad x > -2 (a) State the range of g - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 4

1. To find the inverse function $g^{-1}(x)$, start with the equation:

$y = 3 + \sqrt{x + 2}$.

2. Rearranging this gives:

$\sqrt{x + 2} = y - 3$.

3. Next, square both sides:

$x + 2 = (y - 3)^2$.

4. Then, solve for $x$:

$x = (y - 3)^2 - 2.$
Thus, the inverse function is:

$g^{-1}(x) = (x - 3)^2 - 2.$
5. The domain of $g^{-1}(x)$ is the range of $g(x)$, which is:

Domain: $[3, \infty)$

To find $x$ such that $g(x) = x$, set:

$3 + \sqrt{x + 2} = x.$
Rearranging gives:

$\sqrt{x + 2} = x - 3.$
Squaring both sides yields:

$x + 2 = (x - 3)^2.$
Expanding the right side:

$x + 2 = x^2 - 6x + 9.$
Rearranging results in the quadratic equation:

$x^2 - 7x + 7 = 0.$
Using the quadratic formula:

$x = \frac{7 \pm \sqrt{49 - 28}}{2} = \frac{7 \pm \sqrt{21}}{2}.$
Thus, the solutions are:

$x = \frac{7 + \sqrt{21}}{2}$ and $x = \frac{7 - \sqrt{21}}{2}$.

To find the value of $a$ such that $g(a) = g^{-1}(a)$, we can use the expressions derived earlier:

1. From $g(a)$, we have:

$g(a) = 3 + \sqrt{a + 2}.$
2. From the inverse, we have:

$g^{-1}(a) = (a - 3)^2 - 2.$
Setting these equal:

$3 + \sqrt{a + 2} = (a - 3)^2 - 2.$
Upon solving, we can substitute values of $a$ derived previously to find:

$g(a) = g^{-1}(a), \quad a = \frac{7 + \sqrt{21}}{2}$ or
$a = \frac{7 - \sqrt{21}}{2}.$
Substituting back provides the confirmed value for whether solutions yield consistency.