Determine the range of values of x for which the function f(x) = 2x^3 + 9x^2 - 24x + 6 is strictly decreasing. - Scottish Highers Maths - Question 10 - 2023

Next, we set the derivative equal to zero to find the critical points:

$6x^2 + 18x - 24 = 0$

Dividing the entire equation by 6:

$x^2 + 3x - 4 = 0$

Now, we can factor the quadratic:

$(x + 4)(x - 1) = 0$

The critical points are therefore: $x = -4$ and $x = 1$.

To determine where the function is strictly decreasing, we will analyze the sign of the derivative in the intervals defined by the critical points: $(-\infty, -4)$, $(-4, 1)$, and $(1, \infty)$.

1. Interval $(-\infty, -4)$: Choose $x = -5$: $f'(-5) = 6(-5)^2 + 18(-5) - 24 = 150 - 90 - 24 = 36 > 0$ (increasing)

2. Interval $(-4, 1)$: Choose $x = 0$: $f'(0) = 6(0)^2 + 18(0) - 24 = -24 < 0$ (decreasing)

3. Interval $(1, \infty)$: Choose $x = 2$: $f'(2) = 6(2)^2 + 18(2) - 24 = 24 + 36 - 24 = 36 > 0$ (increasing)

From the intervals analyzed, we can conclude that the function is strictly decreasing on the interval:

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