How to Determine the Concavity of a Function: 4 Clear Steps

A function is concave up on intervals where its second derivative is positive (f’‘(x) > 0) and concave down where its second derivative is negative (f’‘(x) < 0); the points where the sign of f’‘(x) switches — provided f’‘(x) is zero or undefined there — are inflection points where the curve changes its bending direction. To determine the concavity of a function, you take the second derivative, solve f’‘(x) = 0 to find candidate points, test the sign of f’'(x) in each interval, and label each interval concave up or concave down. The entire method reduces to one question: what sign is the second derivative on each interval?
What Concavity Means
Concavity describes which way a curve bends. A function is concave up on an interval if the graph opens upward, like a cup or a smile. A function is concave down on an interval if the graph opens downward, like a frown or an umbrella.
A practical way to picture it: if you draw a chord (straight line) between any two points on a concave-up curve, the curve dips below the chord. On a concave-down curve, the curve bulges above the chord. This “chord test” captures the same idea calculus expresses with derivatives, but visually.
Concavity is separate from whether a function is increasing or decreasing. A curve can be rising while bending downward, or falling while bending upward. Slope tells you direction; concavity tells you how the direction itself is changing. These are the two independent pieces of information a first and second derivative provide together.
Concave up vs. concave down at a glance
| Feature | Concave up | Concave down |
|---|---|---|
| Graph shape | Opens upward (cup) | Opens downward (frown) |
| Second derivative | f’'(x) > 0 | f’'(x) < 0 |
| Chord test | Curve lies below chord | Curve lies above chord |
Find the Second Derivative
The first derivative, f’(x), measures the slope of the curve — how fast the function is rising or falling. The second derivative, f’'(x), measures how the slope itself is changing. That rate of change of slope is exactly what concavity tracks.
Computing f’‘(x) is mechanical: differentiate f(x) once to get f’(x), then differentiate again to get f’'(x).
- For f(x) = x², f’(x) = 2x, and f’'(x) = 2.
- For f(x) = x³, f’(x) = 3x², and f’'(x) = 6x.
Why does the sign of f’’ control concavity? When f’‘(x) is positive, the slope is increasing — the curve is steepening (or lessening its decline), which forces it to bend upward. When f’'(x) is negative, the slope is decreasing, and the curve bends downward. This relationship is a standard result developed in any introductory calculus course; see Khan Academy and Paul’s Online Math Notes for more on the underlying differentiation rules.
Make sure f(x) is twice differentiable on the interval you care about. Points where f’’ is undefined (such as vertical asymptotes or corners) also deserve attention, because concavity can change there too.
Locate Inflection Points
An inflection point is a point on the curve where concavity changes — from up to down, or down to up. Not every point where f’'(x) = 0 is an inflection point; it is only a candidate. The concavity must actually change as you pass through it.
To find candidates, solve f’‘(x) = 0. Then check everywhere f’'(x) is undefined, since these locations can also mark a switch. List all such x-values and use them to split the domain into intervals.
Test for a genuine sign change
Rather than trusting a single value, verify that the sign of f’’ actually flips from one side of the candidate to the other.
- Pick a test number inside the interval just left of the candidate, and evaluate f’’ there.
- Pick a test number inside the interval just right of the candidate, and evaluate f’’ there.
- If the signs differ, concavity changes — you have a real inflection point. If the signs match, the concavity is the same on both sides, so it is not an inflection point.
This sign-change confirmation is the crucial check. For example, f(x) = x⁴ has f’‘(x) = 12x², which is zero at x = 0 but positive on both sides — so x = 0 is not an inflection point despite f’'(0) = 0.
Determine Concavity Intervals
Here is the full four-step process, applied to a worked example. Take f(x) = x³ − 3x² + 1.
Step 1 — Find the second derivative. f’(x) = 3x² − 6x, so f’'(x) = 6x − 6.
Step 2 — Solve f’'(x) = 0 and note any undefined points. Set 6x − 6 = 0, giving x = 1. The second derivative is a polynomial, so it is defined everywhere; x = 1 is the only candidate.
Step 3 — Build intervals and test the sign of f’'. The candidate x = 1 splits the number line into two intervals: (−∞, 1) and (1, ∞).
- Test x = 0 in f’‘(x) = 6x − 6: f’'(0) = −6, which is negative.
- Test x = 2: f’'(2) = 6(2) − 6 = 6, which is positive.
Step 4 — Classify each interval. Because the sign changes from negative to positive across x = 1, the function is concave down on (−∞, 1) and concave up on (1, ∞). The point x = 1 is therefore an inflection point, and its y-value is f(1) = 1³ − 3(1)² + 1 = −1, so the inflection point is (1, −1).
The logic generalizes cleanly. Mark every candidate from Step 2 on a number line, choose a test value in each sub-interval, evaluate the sign of f’’ at that test value, and write “concave up” or “concave down” under each interval. That table of signs is your final answer.
Two checks keep you honest. First, never assume a zero of f’’ is an inflection point until you confirm the sign flip. Second, test with actual numbers rather than eyeballing — for higher-degree polynomials the sign pattern is not always obvious. MIT OpenCourseWare covers this interval-sign method as part of its single-variable calculus materials if you want to practice more examples.
For functions with discontinuities or undefined second derivatives, include those x-values as interval boundaries too, since concavity can change across them even without f’'(x) = 0.
Key Takeaways
- A function is concave up where f’‘(x) > 0 and concave down where f’'(x) < 0.
- Compute f’'(x) by differentiating f(x) twice; its sign is the entire basis for concavity.
- Find candidates by solving f’‘(x) = 0 and identifying where f’’ is undefined.
- A candidate is a true inflection point only if the sign of f’’ actually changes across it.
- Use test values from each interval formed by the candidates to classify concavity interval by interval.
- Concavity is independent of increasing/decreasing behavior — you need the second derivative, not just the first.