Determining Concavity of Functions over Their Domains
Introduced| English |
|---|
| concavity/kənˈkævɪti/ |
| sign of the second derivative/saɪn ɒvðə ˈsekənd dɪˈrɪvətɪv/ |
| concave up/kɒnˈkeɪv ʌp/ |
| concave down/kɒnˈkeɪv daʊn/ |
| point of inflection/pɔɪnt ɒv ɪnˈflekʃn/ |
How the curve bends
- Increasing/decreasing tells you which way the curve goes; concavity 凹凸性 tells you how it bends.
- A curve can rise while bending like a cup (opening up) or like a cap (opening down).
- The bending is controlled by the sign of the second derivative 二阶导数符号.
- $f''$ measures how the slope $f'$ itself is changing.
Concave up: $f'' > 0$
- Where $f''(x)>0$, the graph is concave up 上凹 — shaped like a smile, holding water.
- The slope $f'$ is increasing (getting more positive or less negative).
- Cup-shaped valleys are concave up.
- Tangent lines lie below a concave-up curve.
A graph is concave up exactly where...
Concave up ⇔ $f''>0$ (slope increasing).
Concave up means the slope $f'$ is...
Concave up ⇔ $f'$ increasing ⇔ $f''>0$.
Concave down: $f'' < 0$
- Where $f''(x)<0$, the graph is concave down 下凹 — shaped like a frown.
- The slope $f'$ is decreasing.
- Cap-shaped hills are concave down.
- Tangent lines lie above a concave-down curve.
Where $f''(x)<0$, the graph is concave ____ (cap-shaped).
Negative second derivative → concave down.
Inflection points: where the bend flips
- A point of inflection 拐点 is where concavity changes — from up to down or down to up.
- There, $f''$ changes sign (usually passing through $0$, sometimes undefined).
- Build a sign chart of $f''$, just like you did for $f'$, to find these.
- An inflection point is about the bend changing, not the slope.
Concave down then up
y = x³
This cubic bends like a cap for $x<0$ and a cup for $x>0$ — the flip at $0$ is a point of inflection.
An inflection point requires $f''$ to change sign, not just equal zero.
$y=x^4$ has $f''(0)=0$ but no sign change → not an inflection point.
For $f(x)=x^3$, $f''(x)=6x$. At what $x$ is the inflection point?
$f''=6x$ changes sign at $x=0$.
For $y=x^4$, is $x=0$ an inflection point?
$f''=12x^2\ge0$ on both sides — no sign change, so no inflection.
$f''(c)=0$ alone does not make $c$ an inflection point — $f''$ must actually change sign there. For $y=x^4$, $f''(0)=0$, but $f''\ge0$ on both sides (no sign change), so $x=0$ is not an inflection point. Check the sign on both sides, don't just solve $f''=0$.
Find the concavity and inflection point of $f(x)=x^3$.
- $f'(x)=3x^2$, $f''(x)=6x$.
- $f''<0$ for $x<0$ (concave down); $f''>0$ for $x>0$ (concave up).
- $f''$ changes sign at $x=0$ → inflection point at $(0,0)$.
Concavity is set by the sign of $f''$: $f''>0$ → concave up (cup), $f''<0$ → concave down (cap). A point of inflection is where $f''$ changes sign — the bend flips. $f''=0$ is a candidate, but you must confirm a sign change.