Given, f(x) = x³ – 12x² + 45x + 8. At which point does f(x) has its maximum?

We have, f(x) = x³ ? 12x² + 45x + 8 ???.(1)
Differentiating both sides of (1) with respect to x we
f?(x) = 3x² ? 24x + 45
3x² ? 24x + 45 = 0
Or x² ? 8x + 15 = 0
Or (x ? 3)(x ? 5) = 0
Thus, either x ? 3 = 0 i.e., x = 3 or x ? 5 = 0 i.e., x = 5
Therefore, f?(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f?(3 ? h) = 3*(3 ? h ? 3)(3 ? h ? 5) = 3h(h + 2) = positive.
f?(3 + h) = 3*(3 + h ? 3)(3 + h ? 5) = 3h(h – 2) = negative.
Clearly, f?(x) changes sign from positive on the left to negative on the right of the point x = 3.
So, f(x) has maximum at 3.