Difficulty: Intermediate | Prerequisites: First derivative, critical points, intervals of increase/decrease (Chapter 2 notes)
Concavity describes whether a curve bends upward or downward. You find it by taking the second derivative and checking its sign. Where f''(x) > 0, the graph is concave up (cup-shaped); where f''(x) < 0, it is concave down (cap-shaped). Inflection points are where concavity switches.
Concavity
The direction a curve bends on a given interval. A function's concavity tells you whether the slope itself is increasing or decreasing. Think of it as whether the curve is "opening upward" like a bowl or "opening downward" like an umbrella.
Concave upward (concave up)
A function is concave upward on an interval if f'(x) is increasing on that interval, which means f''(x) > 0. In simple terms, the curve bends upward, shaped like the letter U.
Concave downward (concave down)
A function is concave downward on an interval if f'(x) is decreasing on that interval, which means f''(x) < 0. In simple terms, the curve bends downward, shaped like an upside-down U.
Second derivative, f''(x)
The derivative of the first derivative. It measures the rate of change of the slope. Positive second derivative means the slope is getting steeper (or less negative), negative means the slope is getting less steep (or more negative).
Inflection point
A point where the concavity of f changes from up to down or from down to up. At an inflection point, f''(x) = 0 or f''(x) is undefined, and the sign of f'' actually changes. Think of it as the moment the curve switches from bending one way to bending the other.
Sign chart (for concavity)
A number-line diagram that shows where f''(x) is positive, negative, or zero. You find the zeros of f'', mark them on a number line, then test a value from each interval to determine the sign. This tells you the concavity on each interval.
If f(x) is differentiable on an interval:
The graph is concave upward on (a, b) if f'(x) is increasing on that interval.
The graph is concave downward on (a, b) if f'(x) is decreasing on that interval.
The second derivative test for concavity translates this directly:
f''(x) > 0 on an interval means concave up on that interval.
f''(x) < 0 on an interval means concave down on that interval.
Find f'(x), then find f''(x).
Set f''(x) = 0 and solve. Also note where f''(x) is undefined.
Place these values on a number line, creating test intervals.
Pick a test point in each interval and evaluate the sign of f''(x).
Positive sign means concave up, negative means concave down.
f'(x) = 12x⁵ - 20x³ + 7
f''(x) = 60x⁴ - 60x²
Set f''(x) = 0: 60x⁴ - 60x² = 0, so 60x²(x² - 1) = 0, giving 60x²(x + 1)(x - 1) = 0.
Solutions: x = 0, x = -1, x = 1.
Test intervals with sign chart on (-∞, -1), (-1, 0), (0, 1), (1, ∞):
(-∞, -1): f'' > 0, concave up.
(-1, 0): f'' < 0, concave down.
(0, 1): f'' < 0, concave down.
(1, ∞): f'' > 0, concave up.
Result: concave up on (-∞, -1) ∪ (1, ∞), concave down on (-1, 0) ∪ (0, 1).
f'(x) = 15x⁴ - 20x³
f''(x) = 60x³ - 60x²
Set f''(x) = 0: 60x³ - 60x² = 0, so 60x²(x - 1) = 0.
Solutions: x = 0, x = 1.
Test intervals with sign chart on (-∞, 0), (0, 1), (1, ∞):
(-∞, 0): f'' < 0, concave down.
(0, 1): f'' < 0, concave down.
(1, ∞): f'' > 0, concave up.
Result: concave up on (1, ∞), concave down on (-∞, 0) ∪ (0, 1).
Note: x = 0 is not an inflection point here because the concavity does not change at x = 0 (it stays concave down on both sides).
f'(x) = 2x
f''(x) = 2
Since 2 > 0 always, f(x) = x² is concave up on (-∞, ∞).
There are no inflection points because f'' never equals zero or changes sign.
f''(x) > 0 \Rightarrow \text{concave up}f''(x) < 0 \Rightarrow \text{concave down}\text{Inflection point candidates: solve } f''(x) = 0 \text{ or } f''(x) \text{ undefined}The factoring pattern that appears repeatedly in these problems:
60x^4 - 60x^2 = 60x^2(x^2 - 1) = 60x^2(x+1)(x-1)60x^3 - 60x^2 = 60x^2(x - 1)Factor out the greatest common factor first, then use difference of squares or other factoring techniques to find all zeros.
Concavity shows up whenever you need to know whether a rate of change is accelerating or decelerating. In business, if a revenue function is concave down at a particular time, the revenue is still growing but the growth is slowing, which signals the approach of a peak. Engineers use concavity to determine the stress distribution in beams: a concave-up bending moment diagram tells you the beam is sagging under load.
Students often think that f''(x) = 0 automatically means there is an inflection point. It does not. You must verify that f'' actually changes sign at that value. In Example 2 above, f''(0) = 0 but x = 0 is not an inflection point because the concavity stays concave down on both sides.
Students sometimes confuse "increasing" with "concave up." A function can be increasing and concave down at the same time (picture a hill that is still going up but starting to flatten). Increasing/decreasing is about f'(x), while concavity is about f''(x).
Setting f''(x) = 0 and forgetting to check where f''(x) is undefined. Rational functions can have concavity changes at vertical asymptotes, where f'' does not exist.
Mixing up the sign chart for the first derivative (which gives increasing/decreasing) with the sign chart for the second derivative (which gives concavity). They are separate analyses.
⚠️ "Determine the intervals of concavity" is a standard exam question. You will be expected to find f''(x), factor it, build the sign chart, and state the intervals clearly.
⚠️ Expect at least one problem where f''(x) = 0 at a point that is not an inflection point (like x = 0 in Example 2). This tests whether you check for a sign change.
⚠️ Know the connection between concavity and the second derivative test for local extrema (covered in the next topic). If f'(c) = 0 and f''(c) > 0, then c is a local minimum. If f'(c) = 0 and f''(c) < 0, then c is a local maximum.
⚠️ Be comfortable factoring expressions like 60x⁴ - 60x² quickly. These problems are mechanical once you have the factoring down, but slow if you have to think through the algebra each time.
True or false: if f''(x) > 0 on an interval, the function is concave up on that interval. (True.)
True or false: f''(c) = 0 always means x = c is an inflection point. (False. The sign of f'' must actually change at c.)
Fill in the blank: for f(x) = x², the second derivative is ____, so the function is concave ____ everywhere. (2; up.)
True or false: a function that is decreasing can still be concave up. (True. Think of the right half of a parabola opening upward that has been shifted so it starts above the vertex.)
If f''(x) = 60x²(x - 1), at which x-values might inflection points occur? (x = 0 and x = 1, but only x = 1 is an actual inflection point because f'' does not change sign at x = 0.)
Q: Determine the intervals of concavity for f(x) = x⁴ - 6x² + 5.
A: f'(x) = 4x³ - 12x. f''(x) = 12x² - 12 = 12(x² - 1) = 12(x + 1)(x - 1). Setting f''(x) = 0 gives x = -1 and x = 1. Test intervals: f''(-2) = 12(4 - 1) = 36 > 0 (concave up), f''(0) = 12(0 - 1) = -12 < 0 (concave down), f''(2) = 36 > 0 (concave up). Concave up on (-∞, -1) ∪ (1, ∞), concave down on (-1, 1). Inflection points at x = -1 and x = 1.
Q: For f(x) = x⁵, find f''(x) and determine the concavity intervals.
A: f'(x) = 5x⁴. f''(x) = 20x³. Setting f''(x) = 0 gives x = 0. Test: f''(-1) = -20 < 0 (concave down), f''(1) = 20 > 0 (concave up). Concave down on (-∞, 0), concave up on (0, ∞). Inflection point at x = 0, because f'' changes sign.
Q: If f''(x) = 60x²(x + 1)(x - 1), find the intervals of concavity.
A: f''(x) = 0 at x = 0, x = -1, x = 1. Test intervals: on (-∞, -1), pick x = -2: 60(4)(-1)(-3) = 720 > 0, concave up. On (-1, 0), pick x = -0.5: 60(0.25)(0.5)(-1.5) = -11.25 < 0, concave down. On (0, 1), pick x = 0.5: 60(0.25)(1.5)(-0.5) = -11.25 < 0, concave down. On (1, ∞), pick x = 2: 60(4)(3)(1) = 720 > 0, concave up. Concave up on (-∞, -1) ∪ (1, ∞), concave down on (-1, 0) ∪ (0, 1). Inflection points at x = -1 and x = 1 only. x = 0 is not an inflection point.
Q: Explain why f(x) = x² has no inflection points.
A: f''(x) = 2, which is a positive constant. Because f'' never equals zero and never changes sign, there is no point where concavity switches. The parabola is concave up everywhere.
This connects directly to the second derivative test for local extrema: once you know how to read the sign of f'', you can classify critical points as local maxima or minima without a first derivative sign chart. That test is a shortcut built on concavity.
Concavity also links to curve sketching (Chapter 4 in many textbooks), where you combine information from f, f', and f'' to draw an accurate graph. Knowing the concavity intervals gives you the shape of each piece of the curve.
In business calculus specifically, concavity matters for optimization problems involving revenue, cost, and profit. If your revenue function is concave down at a critical point, that critical point is a maximum, which is exactly the kind of question that appears in applied optimization.
Related Terms / Search Tags
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