Second Derivative Test

AP Calculus · Axiom Academy

Classifying critical points using concavity to determine local maxima and minima The First Derivative Test works by analyzing sign changes of f'(x) around a critical point. The Second Derivative Test offers a shortcut: instead of checking intervals, you evaluate f''(c) at a single point to classify it. Theorem: Second Derivative Test Suppose f'' is continuous near c and f'(c) = 0 . Then: If f''(c) > 0 , then f has a local minimum at c . If f''(c) local maximum at c . If f''(c) = 0 , the test is inconclusive — use the First Derivative Test instead. The logic is straightforward: f''(c) > 0 means the graph is concave up at c , shaping like a bowl that opens upward — so the critical point sits at the bottom (a minimum). Conversely, f''(c) < 0 means concave down, like an upside-down bowl. Classify the critical points of f(x) = x^3 - 3x^2 - 9x + 5 . f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1) Setting f'(x) = 0 : x = 3 and x = -1 . Step 2: Compute the second derivative. Step 3: Evaluate f'' at each critical point. The Second Derivative Test is inconclusive when f''(c) = 0 . Consider f(x) = x^4 : , so x = 0 is a critical point. But for all x , so x = 0 is actually a local (and global) minimum. You must fall back to the First Derivative Test here. Similarly, g(x) = x^3 has g'(0) = 0 and g''(0) = 0 , but x = 0 is neither a max nor a min (it's an inflection point). The second derivative alone can't distinguish these cases. f'(c) = 0 and f''(c) > 0 local minimum

This is the written version of the interactive lesson above. See the full AP Calculus course.