Find the $x$-coordinates of the stationary points on the curve with equation $$y = \frac{1}{2}x^3 - 2x^2 + 6$$ - Scottish Highers Maths - Question 1 - 2019

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

$\frac{3}{2}x^2 - 4x = 0$

Factoring out the common term gives:

$x(\frac{3}{2}x - 4) = 0$

Setting each factor to zero:

1. First factor:

   $x = 0$

2. Second factor:

   $\frac{3}{2}x - 4 = 0$

   Solving for $x$ gives:

   $\frac{3}{2}x = 4$

   $x = \frac{4 \cdot 2}{3} = \frac{8}{3}$

Thus, the $x$-coordinates of the stationary points are:

$x = 0 \text{ and } x = \frac{8}{3}$