Concavity and Inflection Points

Concavity measures the way a curve bends. If the second derivative is positive, the curve is concave up (smiles up like a bowl). If $f''$ is negative, it's concave down (frowns). The points where concavity flips are called inflection points — and they're exactly where $f''(x) = 0$ (provided the sign actually changes).

The Physics — Wait, Just Math

For $f(x) = x^{4} - 4x^{3}$, take two derivatives. Find where $f''(x) = 0$. Test the sign of $f''$ on each side to confirm a real flip in concavity.

Step-by-Step Solution

Given: $f(x) = x^{4} - 4x^{3}$.

Find: The concavity intervals and the inflection points.

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Step 1 — Compute the first derivative.

$f'(x) = 4x^{3} - 12x^{2}$

Step 2 — Compute the second derivative.

$f''(x) = 12x^{2} - 24x$

Step 3 — Factor $f''(x)$ and find its zeros.

$f''(x) = 12x(x - 2)$

Setting $f''(x) = 0$:

$12x(x - 2) = 0 \;\;\Longrightarrow\;\; x = 0 \text{ or } x = 2$

These are the candidates for inflection points. We still need to confirm the concavity actually changes sign at each.

Step 4 — Test the sign of $f''$ in each interval.

The candidates split the line into three intervals. Pick test points:

• At $x = -1$ (in $(-\infty, 0)$): $f''(-1) = 12(-1)(-3) = +36 > 0$ → concave up
• At $x = 1$ (in $(0, 2)$): $f''(1) = 12(1)(-1) = -12 < 0$ → concave down
• At $x = 3$ (in $(2, \infty)$): $f''(3) = 12(3)(1) = +36 > 0$ → concave up

The sign does flip at both $x = 0$ and $x = 2$, so both are genuine inflection points.

Step 5 — Compute $f$ at each inflection point.

$f(0) = 0^{4} - 4(0)^{3} = 0$

$f(2) = 2^{4} - 4(2)^{3} = 16 - 32 = -16$

So the inflection points are at $(0, 0)$ and $(2, -16)$.

Step 6 — (Bonus) Find the critical points using $f'(x)$.

$f'(x) = 4x^{3} - 12x^{2} = 4x^{2}(x - 3) = 0 \;\;\Longrightarrow\;\; x = 0 \text{ or } x = 3$

At $x = 0$: $f''(0) = 0$ → second-derivative test inconclusive — but we already know it's an inflection point, not an extremum.

At $x = 3$: $f''(3) = +36 > 0$ → local minimum with $f(3) = 81 - 108 = -27$.

So $(3, -27)$ is the only local extremum — a global minimum on the visible window.

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Answer:

• $f$ is concave up on $\boxed{(-\infty,\, 0) \cup (2,\, \infty)}$
• $f$ is concave down on $\boxed{(0,\, 2)}$
• Inflection points at $(0, 0)$ and $(2, -16)$
• The only local extremum is the local minimum at $(3, -27)$

Try It

• Slide the point along the curve — green stars mark the inflection points.
• The curve is drawn in cyan where concave up (smiling) and purple where concave down (frowning).
• Toggle show $f''(x)$ to see the second derivative — it crosses zero exactly at the inflection points.
• The HUD lights up with "↑ CONCAVE UP" or "↓ CONCAVE DOWN" depending on the sign of $f''$ at your current point.