Functions $f$ and $g$ are defined on suitable domains by $f(x) = 10 + x$ and $g(x) = (1 + x)(3 - x) + 2$ - Scottish Highers Maths - Question 2 - 2015

From our previous answer, we have:

$f(g(x)) = 15 + 2x - x^2$

To express this in the form $p(x^2 + qx) + r$, identify coefficients:

1. Rearranging gives: $-x^2 + 2x + 15 = -1(x^2 - 2x) + 15$

2. Hence:

   • Let $p = -1$, $q = -2$, and $r = 15$.

Thus, $f(g(x)) = -1(x^2 - 2x) + 15$

The function $h(x)$ is given by:

$h(x) = \frac{1}{f(g(x))}$

For $h(x)$ to be defined, the denominator $f(g(x))$ cannot be zero:

1. Set $f(g(x)) = 0$: $15 + 2x - x^2 = 0$

2. Rearranging gives: $x^2 - 2x - 15 = 0$

3. Solving this quadratic using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2(1)}$ $= \frac{2 \pm \sqrt{4 + 60}}{2} = \frac{2 \pm \sqrt{64}}{2}$ $= \frac{2 \pm 8}{2}$

4. The solutions are: $x = 5 \quad \text{and} \quad x = -3$

Therefore, the values of $x$ that cannot be in the domain of $h$ are: $x = 5 \quad \text{and} \quad x = -3$