Q33: Absolute maximum and minimum of $2x^3 - 15x^2 + 36x + 1$ on $[1, 5]$

Differentiate term by term with the power rule $\frac{d}{dx}x^n = nx^{n - 1}$⁠: $2\cdot3x^2 = 6x^2$⁠, $15\cdot2x = 30x$⁠, and the constant $1$ gives $0$⁠.

Take out the common factor $6$⁠. Then factorise $x^2 - 5x + 6$⁠: we need two numbers with sum $5$ and product $6$⁠, which are $2$ and $3$⁠.

The critical points are where $f'(x) = 0$⁠. ($f$ is a polynomial, so $f'(x)$ exists everywhere and there are no other critical points.)

$6(x - 2)(x - 3) = 0$ when one factor is 0: $x - 2 = 0$ gives $x = 2$⁠, and $x - 3 = 0$ gives $x = 3$⁠.

Keep only the critical points inside the given interval. Both $2$ and $3$ lie in $[1, 5]$⁠.

This is the method for a closed interval. A continuous function on $[a, b]$ always has an absolute maximum and an absolute minimum, and each one occurs either at a critical point or at an end point $a$ or $b$⁠.

So we only need $f$ at $x = 1, 2, 3, 5$⁠, and then we compare. No second derivative test is needed.

Substitute carefully, term by term. For example, at $x = 5$⁠:

All four, each term first and then the running sum:

Watch out: $f(2) = 29$ is only a local maximum and $f(3) = 28$ only a local minimum. The end points give larger and smaller values, so the absolute values come from the end points here.