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

To find the inverse function, start with:

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

1. Rearranging gives: [ \sqrt{x + 2} = y - 3 ]
2. Squaring both sides results in: [ x + 2 = (y - 3)^2 ]
3. Thus, [ x = (y - 3)^2 - 2 ]
4. Therefore, the inverse function is: [ g^{-1}(x) = (x - 3)^2 - 2 ]
5. The domain for $g^{-1}(x)$ matches the range of $g(x)$, which is $[3, \infty)$.

To find the value of $x$ such that:

$g(x) = x$

1. Setting up the equation gives: [ 3 + \sqrt{x + 2} = x ]
2. Rearranging leads to: [ \sqrt{x + 2} = x - 3 ]
3. Squaring both sides gives: [ x + 2 = (x - 3)^2 ]
4. Expanding the right-hand side: [ x + 2 = x^2 - 6x + 9 ]
5. Rearranging yields: [ x^2 - 7x + 7 = 0 ]
6. Using the quadratic formula, we find: [ x = \frac{7 \pm \sqrt{49 - 28}}{2} = \frac{7 \pm \sqrt{21}}{2} ]

From part (c), the solutions for $x$ can be either: [ a_1 = \frac{7 + \sqrt{21}}{2} ] or [ a_2 = \frac{7 - \sqrt{21}}{2} ]

1. We can substitute $a$ into either function $g(a)$ or $g^{-1}(a)$ and equate them.
2. Thus, the values of $a$ must correspond to both parts $g(a) = g^{-1}(a)$, leading to: [ g(a) = 3 + \sqrt{a + 2} ] for values of $a$ calculated alongside the previous findings.