Math · Calculus I · Worked example
When the second derivative test fails
Find the local extrema and inflection points of f(x) = x⁴ − 4x³.
Find the critical points
f′(x) = 4x³ − 12x² = 4x²(x − 3), which is zero at x = 0 and x = 3.
Try the second derivative test
f″(x) = 12x² − 24x. At x = 3, f″(3) = 36 > 0, so f(3) = 81 − 108 = −27 is a local minimum. At x = 0, f″(0) = 0 and the test gives no answer.
Use the first derivative test at x = 0
The factor 4x² is never negative, so f′ has the sign of x − 3, which is negative on both sides of 0. The graph falls through x = 0 without turning, so there is no extremum there.
| Interval | Test x | f′(x) | f is |
|---|---|---|---|
| x < 0 | −1 | −16 | decreasing |
| 0 < x < 3 | 1 | −8 | decreasing |
| x > 3 | 4 | 64 | increasing |
Find the inflection points
f″(x) = 12x(x − 2) changes sign at x = 0 and x = 2: concave up for x < 0, down for 0 < x < 2 and up for x > 2. Both (0, 0) and (2, −16) are inflection points.
Result
The only local extremum is the minimum f(3) = −27. The inflection points are (0, 0) and (2, −16); x = 0 is a critical point but not an extremum.
Your turn
Does f(x) = x³ have a local extremum at x = 0?
Show the answer and explanation
No. f′(0) = 0, but f keeps increasing through 0, and (0, 0) is an inflection point.
f′(x) = 3x² is positive on both sides of 0, so f′ does not change sign and there is no maximum or minimum. f″(x) = 6x changes sign at 0, so the concavity changes there.
Keep exploring
Open the Derivative Tracer at x = 0: the tangent is flat, yet the curve keeps falling on both sides.
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