Curve Sketching: Increasing/Decreasing & Concavity
The first derivative reveals where a function increases/decreases and locates local extrema; the second derivative reveals concavity and inflection points.
First derivative test
Where f'(x) > 0, f is increasing; where f'(x) < 0, f is decreasing. At a critical point, if f' changes from + to -, that's a local max; from - to +, a local min; if f' doesn't change sign, it's neither.
Second derivative test and concavity
Where f''(x) > 0, f is concave up (curve opens upward, like a cup — tangent lines lie below the curve). Where f''(x) < 0, f is concave down (tangent lines lie above the curve).
An inflection point is where concavity changes (f'' changes sign), not merely where f''=0 — always verify the sign change, just like with critical points.
Second derivative test (alternative to first derivative test at critical points): if f'(c)=0 and f''(c)>0, c is a local min; if f''(c)<0, c is a local max. If f''(c)=0, the test is inconclusive — fall back to the first derivative test.
Worked Example
For f(x) = x³ - 3x², find intervals of concavity and any inflection point.
- f'(x) = 3x² - 6x. f''(x) = 6x - 6.
- Set f''(x)=0: 6x-6=0 → x=1.
- Test signs: for x<1, f''<0 (concave down); for x>1, f''>0 (concave up). Sign changes at x=1, confirming an inflection point.
Answer: Concave down on (-∞,1), concave up on (1,∞); inflection point at x=1.
Key terms (2)
- Inflection point
- A point where concavity changes — where f'' changes sign.
- Concave up / down
- f''>0 means concave up (like a cup); f''<0 means concave down (like a cap).
Practice Quiz
f'(x) changes from negative to positive at x=3. What happens at x=3?