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 $g.$ (b) Find $g^{-1}(x)$ and state its domain - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 4

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

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

To solve for $x$, rearranging gives:

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

Squaring both sides results in:

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

Solving for $x$ gives:

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

Thus, the inverse function can be expressed as:

$g^{-1}(x) = (x - 3)^2 - 2.$

For the domain, we need $g(x)$ to be defined, which occurs for $x > -2$, hence the domain of $g^{-1}(x)$ is:

$[3, \infty).$

To find the exact value of $x$ where $g(x) = x$, we set:

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

Rearranging yields:

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

Squaring both sides gives:

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

Expanding the right side and rearranging leads to:

$x + 2 = x^2 - 6x + 9$

This simplifies to:

$x^2 - 7x + 7 = 0.$

Applying the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{49 - 28}}{2} = \frac{7 \pm \sqrt{21}}{2}.$

Thus, the exact values of $x$ are:

$x = \frac{7 + \sqrt{21}}{2} \quad \text{or} \quad x = \frac{7 - \sqrt{21}}{2}.$

We need to find the value of $a$ such that:

$g(a) = g^{-1}(a).$

Using the expression for $g(x)$ and $g^{-1}(x)$, we already derived:

$g(a) = 3 + \sqrt{a + 2}, \quad g^{-1}(a) = (a - 3)^2 - 2.$

Equating gives:

$3 + \sqrt{a + 2} = (a - 3)^2 - 2.$

Solving this can be complex but, we can substitute the values we found for $x$. Testing:

$a = \frac{7 + \sqrt{21}}{2}, \quad a = \frac{7 - \sqrt{21}}{2}$

Would yield the values for $a$, but generally, you would simplify further based on context.