Let f(x) = x⁴ - 4x³. A student finds the critical points and tries the second derivative test at…

Step 1. The critical points. $f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$, so $x = 0$ and $x = 3$.

Step 2. The second derivative. $f''(x) = 12x^2 - 24x$. At $x = 0$, $f''(0) = 0$, so the test fails. At $x = 3$, $f''(3) = 108 - 72 = 36 > 0$, so there is a local minimum.

Step 3. The point $x = 0$. $f'(x) = 4x^2(x - 3)$. For small $x$ on either side of $0$, $x^2 > 0$ and $x - 3 < 0$, so $f' < 0$. The sign does not change. So $x = 0$ is neither a maximum nor a minimum.

Step 4. The minimum value. $f(3) = 81 - 108 = -27$.

Step 5. The maxima. The only extreme point is the minimum at $x = 3$, so there are $0$ local maxima.