A function, f, is defined on ℝ, the set of real numbers, by f(x) = x³ + 3x² - 9x + 5 - Scottish Highers Maths - Question 8 - 2023

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

$3x^2 + 6x - 9 = 0$

We can simplify this equation by dividing the entire equation by 3:

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

Now we can factor or use the quadratic formula to solve for x. Factoring gives:

$(x + 3)(x - 1) = 0$

Thus, the solutions are:

$x = -3 \text{ and } x = 1$

To find the y-coordinates for these x-values, substitute them back into the original function:

For $x = -3$: $f(-3) = (-3)^3 + 3(-3)^2 - 9(-3) + 5 = -27 + 27 + 27 + 5 = 32$

For $x = 1$: $f(1) = (1)^3 + 3(1)^2 - 9(1) + 5 = 1 + 3 - 9 + 5 = 0$

Thus, the stationary points are:

$(-3, 32) \text{ and } (1, 0)$

To determine the nature of the stationary points, we compute the second derivative:

$f''(x) = 6x + 6$

Now evaluate the second derivative at the stationary points:

At $x = -3$: $f''(-3) = 6(-3) + 6 = -18 + 6 = -12 \text{ (concave down, local maximum)}$

At $x = 1$: $f''(1) = 6(1) + 6 = 6 + 6 = 12 \text{ (concave up, local minimum)}$

In conclusion, the nature of the stationary points are:

1. Local maximum at $(-3, 32)$
2. Local minimum at $(1, 0)$