Functions $f$ and $g$ are given by $f(x) = x^2 - 2$ and $g(x) = 3x + 5$, for $x \in \mathbb{R}$ - Scottish Highers Maths - Question 5 - 2022

To find $g(f(x))$, we substitute $f(x)$ into the function $g$.

Since $f(x) = x^2 - 2$, we have:

$g(f(x)) = g(x^2 - 2) = 3(x^2 - 2) + 5.$

Now, we simplify this:

$g(f(x)) = 3x^2 - 6 + 5 = 3x^2 - 1.$

We need to solve the inequality:

$f(g(x)) < g(f(x)).$

Substituting our earlier results:

$9x^2 + 30x + 23 < 3x^2 - 1.$

Rearranging gives:

$6x^2 + 30x + 24 < 0.$

Factoring out the common term:

$3(2x^2 + 10x + 8) < 0.$

Next, we find the roots of the quadratic equation $2x^2 + 10x + 8 = 0$ using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-10 \pm \sqrt{100 - 64}}{4} = \frac{-10 \pm 6}{4}.$

This gives us the roots:

$x = -1 \text{ and } x = -4.$

The quadratic opens upwards (as the coefficient of $x^2$ is positive). Thus, the expression is negative between the roots:

$-4 < x < -1.$

Therefore, the range of values of $x$ is:

$$\boxed{(-4, -1)}.$