Concavity and the Second Derivative
"The curve is flattening." "Growth is slowing." "Cases are still rising, but more slowly than
last week." Every one of those headlines — from pandemics to profits to house prices — is a
claim about the same mathematical object: not the value of a quantity, not even its
rate of change, but the rate of change of the rate of change.
Picture two companies whose revenues are both climbing. One's growth is accelerating — every
quarter adds more than the last; its graph bows upward like the inside of a skateboard ramp.
The other is still growing, but each quarter adds a little less; its graph arches over like a
hill just before its summit. Same direction, opposite bend. An investor who can read
that bend sees the future of the trend, not just its present. This bending is called
concavity, and it is what the second derivative measures.
Cup or dome: the sign of f''
Two curves can both be rising, yet curve in opposite ways: one bows up like a cup, the other
arches over like a dome. The
second derivative
f''(x) tells them apart:
- f''(x) > 0: concave up \smile — the curve holds water; the slope is increasing.
- f''(x) < 0: concave down \frown — the curve spills water; the slope is decreasing.
Concavity is just "the slope of the slope". f''>0 means
f' is going up, so the tangents tilt steeper and steeper as you walk
right — even if they start negative. Read that sentence again, because it is the whole idea in
one line: concave up means the slopes are increasing; it says nothing about whether the
curve itself is going up.
The same idea in three languages:
- Words: concave up = the curve turns steadily anticlockwise as you travel along it (slopes rise); concave down = it turns clockwise (slopes fall).
- Symbols: concave up on I exactly when f''(x) > 0 throughout I.
- Pictures: on a concave-up stretch every tangent line lies below the curve; on a concave-down stretch every tangent lies above it. Watch for this in the interactive figure next.
Feel the bend
Below is f(x) = x^3 - 3x (bold) with
f''(x) = 6x (thin, dashed). Slide the tangent along the curve and
watch two things at once. First, the tangent's tilt: on the left it gets shallower and
shallower (slope decreasing, f''<0, a frown), then past the middle it
steepens again (slope increasing, f''>0, a smile). Second, which
side of the curve the tangent sits on: above the curve on the concave-down half, below
it on the concave-up half.
And check the thin line as you slide: the moment the tangent stops lying above the curve and
starts lying below is exactly the moment f''(x)=6x crosses zero.
Worked example 1 — concavity intervals from f''
Suppose f'' exists on an interval I.
- If f''(x) > 0 for all x in I, then f is concave up on I.
- If f''(x) < 0 for all x in I, then f is concave down on I.
Problem. Find the intervals on which
f(x) = x^4 - 6x^2 is concave up and concave down.
Step 1 — differentiate twice.
f'(x) = 4x^3 - 12x, \qquad f''(x) = 12x^2 - 12 = 12(x-1)(x+1).
Step 2 — find where f'' could change sign.
f''(x) = 0 at x = -1 and
x = 1. These two points chop the number line into three pieces.
Step 3 — test the sign of f'' on each piece. Pick
an easy sample point in each:
f''(-2) = 36 > 0, \qquad f''(0) = -12 < 0, \qquad f''(2) = 36 > 0.
Step 4 — conclude. f is concave up on
(-\infty, -1) and (1, \infty), and
concave down on (-1, 1). The bend flips at
x = \pm 1, and since f(\pm 1) = -5, the
curve has inflection points at (\pm 1, -5) — see them in the figure:
Inflection points — and Worked example 2
An inflection point is where the concavity switches — the curve
changes from cup to dome or back. There f''(x)=0 (or is undefined)
and actually changes sign. For f(x)=x^3-3x:
f''(x) = 6x = 0 \;\Rightarrow\; x = 0, \quad f'' < 0 \text{ left of } 0,\; f'' > 0 \text{ right} \;\Rightarrow\; \text{inflection at } x = 0
Just like with extrema, f''=0 is only a candidate — you
must check the sign really flips. Here is the full ritual on a slightly meatier cubic.
Problem. Locate the inflection point of
g(x) = x^3 - 6x^2 + 9x + 1.
Step 1 — differentiate twice.
g'(x) = 3x^2 - 12x + 9, \qquad g''(x) = 6x - 12.
Step 2 — find the candidate.
g''(x) = 0 \Rightarrow 6x - 12 = 0 \Rightarrow x = 2.
Step 3 — confirm the sign change.
g''(1) = -6 < 0 (concave down to the left) and
g''(3) = 6 > 0 (concave up to the right). The sign genuinely flips,
so x = 2 is a real inflection point.
Step 4 — give the point, not just the x.
g(2) = 8 - 24 + 18 + 1 = 3, so the inflection point is
(2, 3). (An exam question asking for the point wants both
coordinates — a classic mark lost.)
Two traps catch almost every calculus student at least once:
-
f''(c) = 0 does NOT guarantee an inflection point.
Take f(x) = x^4: then f''(x) = 12x^2,
so f''(0) = 0 — a perfect candidate. But
12x^2 \ge 0 on both sides of zero: the sign never
changes, the cup never flips, and x=0 is a minimum, not an
inflection. The zero of f'' only nominates a candidate; the
sign change elects it.
-
Concave up does not mean increasing. The left half of
y = x^2 is falling steeply — yet it is concave up the whole way,
because its slope is increasing (from very negative towards zero). Direction is
f''s business; bend is f'''s. A curve
can be any combination of the two.
The second-derivative test
Concavity gives a quick shortcut for classifying a critical point where
f'(c)=0: just check the bend there. A flat spot at the bottom of a
cup must be a valley; a flat spot on top of a dome must be a peak.
- f''(c) > 0 (concave up, a cup): local minimum.
- f''(c) < 0 (concave down, a dome): local maximum.
- f''(c) = 0: inconclusive — fall back to the first-derivative test.
For f(x)=x^3-3x at x=1:
f''(1)=6>0, concave up, so it's a minimum — matching what the
sign line told us. Step the diagram to see both critical points classified by their bend.
Worked example 3 — reading a story from the bend
s(t) is a car's position along a road. Then
s' is its velocity and s'' its
acceleration — so the concavity of the position graph literally is the pressure on
the pedals. Suppose at some instant s'(t) > 0 and
s''(t) < 0. What is the car doing?
Step 1 — read s'. Velocity positive: the car is
moving forward.
Step 2 — read s''. Acceleration negative: the
velocity is decreasing. Moving forward + slowing down = the brakes are on.
Step 3 — see it on the graph. The position graph is rising (going forward)
but concave down (flattening out) — the exact shape of a car coasting up to a red light. The
four sign combinations tell four different stories:
- s' > 0,\ s'' > 0: forward and speeding up (foot on the gas).
- s' > 0,\ s'' < 0: forward but braking.
- s' < 0,\ s'' < 0: reversing and speeding up in reverse.
- s' < 0,\ s'' > 0: reversing but slowing — about to stop or turn around.
The same dictionary works for any quantity: population, revenue, temperature, infections.
"Still rising but concave down" is always the phrase for the peak is coming.
In spring 2020 the phrase "flattening the curve" went from epidemiology
seminars to every front page on Earth. The curve was cumulative infections, and the day
everyone was desperate to reach was precisely an inflection point: while the
graph is concave up, each day brings more new cases than the day before; the moment
it turns concave down, each day brings fewer — the epidemic is still growing, but the
tide has turned. Epidemiologists watch f'', not
f.
Economists talk the same way — "the recovery has reached an inflection point", "inflation is
decelerating". Which invites the most famous third-derivative joke in history: in 1972 US
President Nixon announced that the rate of increase of inflation was decreasing.
Prices p: inflation is p', its increase
is p'', and that increase decreasing is
p''' < 0 — quite possibly the first time a head of state used the
third derivative to claim things were going well. (Prices were still accelerating upward.
Always check which derivative a headline is really about.)
Watch it on Khan Academy