Given, f(x) = x³ – 12x² + 45x + 8. What is minimum value of f(x)?

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?(5 ? h) = 3*(5 ? h ? 3)(5 ? h ? 5) = -3h(2 ? h) = negative.
f?(5 + h) = 3*(5 + h ? 3)(5 + h ? 5) = 3h(2 + h) = positive.
Clearly, f?(x) changes sign from negative on the left to positive on the right of the point x = 5.
So, f(x) has minimum at 5.
Putting, x = 5 in (1)
Thus, its maximum value is,
f(3) = 5³ ? 12*5² + 45*5 + 8 = 58.