Applications of derivatives
Set the first derivative to zero for critical points, then let the sign of the second derivative decide maximum or minimum.
A worked example
For f(x) = x³ - 12x² + 45x, f′(x) = 3(x - 3)(x - 5) is zero at x = 3 and x = 5, and f″(5) = 6 > 0, so x = 5 is a local minimum.
The mistake people actually make
A point of inflection is where f″ changes sign, not where f′ is zero. In a related-rates question, differentiate the formula with respect to time before substituting the given value.