Why is x = 0 not the absolute maximum even though it is a local maximum?

f(0) = 1 is the highest point near x = 0, but the right endpoint gives f(4) = 17, which is larger. On a closed interval the endpoints must always be compared with the critical values.

Do I need the second derivative test here?

No. For absolute extrema on a closed interval, just list the values of f at the critical numbers and at the endpoints and pick the largest and smallest. Classifying each point is extra work that does not change the answer.

Calculus 1: Absolute Max and Min of f(x) = x³ − 3x² + 1 on [−1/2, 4] (Closed Interval Method)

The Problem

Find the absolute maximum and absolute minimum values of $f(x) = x^3 - 3x^2 + 1$ on the interval $\left[-\tfrac{1}{2},\ 4\right]$.

This is the textbook example of the Closed Interval Method in a first calculus course, and midterms reuse it with small changes to the interval or the constant.

Why the Method Works

$f$ is a polynomial, so it is continuous on the closed interval $[-\tfrac12, 4]$. By the Extreme Value Theorem it must attain an absolute maximum and an absolute minimum there, and each one happens either at a critical number inside the interval or at an endpoint.

Step 1: Differentiate

$$ f'(x) = 3x^2 - 6x = 3x(x - 2) $$

Step 2: Find Critical Numbers

$f'(x) = 0 \implies x = 0$ or $x = 2$. Since $f'$ is a polynomial, it is never undefined, so there are no other critical numbers. Both 0 and 2 lie inside $[-\tfrac12, 4]$, so keep both.

Step 3: Evaluate f at the Critical Numbers and the Endpoints

$$ f\left(-\tfrac12\right) = -\tfrac18 - \tfrac34 + 1 = \tfrac18 $$

$$ f(0) = 0 - 0 + 1 = 1 $$

$$ f(2) = 8 - 12 + 1 = -3 $$

$$ f(4) = 64 - 48 + 1 = 17 $$

x	Type	f(x)
−1/2	left endpoint	1/8
0	critical number	1
2	critical number	−3
4	right endpoint	17

Step 4: Compare

Absolute maximum: 17, at $x = 4$
Absolute minimum: −3, at $x = 2$

Note the wording: the maximum value is 17; it occurs at $x = 4$. Many graders take points off if you answer "the maximum is 4".

Traps That Cost Points

Skipping the endpoints. The biggest value here is at an endpoint. Students who only test critical numbers report max = 1, which is wrong.
Arithmetic at x = −1/2. $(-\tfrac12)^3 = -\tfrac18$ and $-3(-\tfrac12)^2 = -\tfrac34$. The sum with 1 is $\tfrac18$, not $\tfrac38$ (a sign slip on the cube) and not $-\tfrac78$ (forgetting the $+1$).
Keeping critical numbers outside the interval. On a different interval, such as $[3, 5]$, both 0 and 2 would be discarded and only the endpoints would matter.
Mixing up local and absolute. $x = 0$ is a local maximum and $x = 2$ is a local minimum, but only the comparison table decides the absolute extrema.

Quick Check With a Graph

Graph $y = x^3 - 3x^2 + 1$ in Desmos and restrict the domain with $\{-0.5 \le x \le 4\}$. The lowest point on the curve is the turning point $(2, -3)$ and the highest point is the right end $(4, 17)$, matching the table.

The same "derivative equals zero, then compare" logic drives optimization word problems; see minimizing the surface area of an open box.