Explanation:
f(x) = x³ − 6x² + 9x + 3
Take the derivative and evaluate at x = 2.
f'(x) = 3x² − 12x + 9
f'(2) = -3
Check for local minimums or maximums by setting f'(x) equal to 0.
0 = 3x² − 12x + 9
0 = x² − 4x + 3
0 = (x − 1) (x − 3)
x = 1 or 3
Evaluate f(x) at the critical values, and at the end points.
f(0) = 3
f(1) = 7
f(3) = 3
f(5) = 23
f(x) has a minimum of 3 and a maximum of 23.