Mathematics Extension 1 2018 HIGHER SCHOOL CERTIFICATE EXAMINATION Question: Given the following function, determine its critical points by finding its first derivative: $f(x) = 3x^3 - 6x^2 + 2x + 1$ - HSC - SSCE Mathematics Extension 1 - Question 1 - 2018 - Paper 1

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

$9x^2 - 12x + 2 = 0.$

Using the quadratic formula, where (a = 9), (b = -12), and (c = 2):

$x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 9 \cdot 2}}{2 \cdot 9} = \frac{12 \pm \sqrt{144 - 72}}{18} = \frac{12 \pm \sqrt{72}}{18} = \frac{12 \pm 6\sqrt{2}}{18} = \frac{2 \pm \sqrt{2}}{3}.$

Thus, the critical points are (x = \frac{2 + \sqrt{2}}{3}) and (x = \frac{2 - \sqrt{2}}{3}.)

To classify the critical points, compute the second derivative:

$f''(x) = \frac{d}{dx}(9x^2 - 12x + 2) = 18x - 12.$

Evaluate the second derivative at each critical point:

1. For (x = \frac{2 + \sqrt{2}}{3}):

   $f''\left(\frac{2 + \sqrt{2}}{3}\right) = 18\left(\frac{2 + \sqrt{2}}{3}\right) - 12 = 6(2 + \sqrt{2}) - 12 = 12 + 6\sqrt{2} - 12 = 6\sqrt{2} > 0.$

   Since (f'' > 0), this point is a local minimum.

2. For (x = \frac{2 - \sqrt{2}}{3}):

   $f''\left(\frac{2 - \sqrt{2}}{3}\right) = 18\left(\frac{2 - \sqrt{2}}{3}\right) - 12 = 6(2 - \sqrt{2}) - 12 = 12 - 6\sqrt{2} - 12 = -6\sqrt{2} < 0.$

   Since (f'' < 0), this point is a local maximum.