Critical Points and Extrema of a Cubic

A function's local maxima and minima occur where the derivative is zero (a horizontal tangent) or undefined. These are called critical points. For a smooth polynomial, finding extrema reduces to solving $f'(x) = 0$ and then classifying each root.

The Physics — Wait, Just the Math

For $f(x) = x^{3} - 3x^{2} + 2$, we want the points where the slope vanishes. Once we find them, we use either the first-derivative sign test or the second-derivative test to decide if each is a max, a min, or neither (a "saddle" inflection point).

Step-by-Step Solution

Given: $f(x) = x^{3} - 3x^{2} + 2$.

Find: All critical points and classify them as local max, local min, or neither.

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Step 1 — Compute the first derivative.

Apply the power rule term by term:

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

Step 2 — Set $f'(x) = 0$ and solve.

$3x^{2} - 6x = 0$

Factor:

$3x(x - 2) = 0$

Two roots:

$x = 0 \quad\text{or}\quad x = 2$

These are the two critical points.

Step 3 — Compute $f$ at each critical point.

$f(0) = (0)^{3} - 3(0)^{2} + 2 = 2$

$f(2) = (2)^{3} - 3(2)^{2} + 2 = 8 - 12 + 2 = -2$

So the critical points are $(0, 2)$ and $(2, -2)$.

Step 4 — Classify each using the second-derivative test.

Compute $f''(x)$:

$f''(x) = 6x - 6$

Evaluate at each critical point:

$f''(0) = 6(0) - 6 = -6 < 0 \;\;\Longrightarrow\;\; \text{local maximum at } (0, 2)$

$f''(2) = 6(2) - 6 = +6 > 0 \;\;\Longrightarrow\;\; \text{local minimum at } (2, -2)$

Step 5 — Sanity check with the first-derivative sign test.

Plug test points into $f'(x) = 3x(x - 2)$:

• At $x = -1$: $f'(-1) = 3(-1)(-3) = +9 > 0$ → $f$ increasing
• At $x = 1$: $f'(1) = 3(1)(-1) = -3 < 0$ → $f$ decreasing
• At $x = 3$: $f'(3) = 3(3)(1) = +9 > 0$ → $f$ increasing

So $f$ goes up → down → up, confirming a local max at $x = 0$ and a local min at $x = 2$. ✓

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Answer: The function $f(x) = x^{3} - 3x^{2} + 2$ has two critical points:

• A local maximum at $\boxed{(0,\, 2)}$
• A local minimum at $\boxed{(2,\, -2)}$

There are no other extrema. The function is increasing on $(-\infty, 0)$, decreasing on $(0, 2)$, and increasing again on $(2, \infty)$.

Try It

• Slide the point along the curve — see the tangent line rotate. At $x = 0$ and $x = 2$ it goes flat (horizontal).
• The two critical points are marked with green stars.
• Toggle the show $f'(x)$ option to see the derivative curve below — it crosses zero exactly at $x = 0$ and $x = 2$.
• The HUD shows $f(x)$, $f'(x)$, and $f''(x)$ live, and lights up with "MAX" or "MIN" when you're near a critical point.