9. (a) Find the x-coordinates of the stationary points on the graph with equation y = f(x), where f(x) = x^3 + 3x^2 - 24x - Scottish Highers Maths - Question 9 - 2016

The function f is strictly increasing where its derivative is positive. We analyze the sign of the derivative:

$f'(x) = 3x^2 + 6x - 24.$

From the factored form, we know the critical points (stationary points) are at x = -4 and x = 2. We test intervals based on these points:

1. For the interval $(- ext{∞}, -4)$, choose x = -5:

   • $f'(-5) = 3(-5)^2 + 6(-5) - 24 = 75 - 30 - 24 = 21 > 0$ (increasing)

2. For the interval $(-4, 2)$, choose x = 0:

   • $f'(0) = 3(0)^2 + 6(0) - 24 = -24 < 0$ (decreasing)

3. For the interval $(2, + ext{∞})$, choose x = 3:

   • $f'(3) = 3(3)^2 + 6(3) - 24 = 27 + 18 - 24 = 21 > 0$ (increasing)

Therefore, the function f is strictly increasing for:

• $x < -4$ and $x > 2$.