Where does the concavity of f change?

How it works

f is concave up where f″(x) > 0 and concave down where f″(x) < 0. An inflection point is a point where f is continuous and f″ changes sign.

1. A computer algebra system (nerdamer, open source) finds f′ and then f″ as the derivative of f′, each checked against a numeric difference quotient at 20 points (to 1 part in a million). The algebra runs in the background after you start typing.
2. The zeros and breaks of g = f″ are found between −∞ and ∞, as described below. Call them the cuts. A point where f is not defined is a break of f″ too, so it is a cut. When f″ holds an |u| (|x| − x² gives 1/|x| − 2 − x²/|x|³), f″ can jump where u = 0 without changing sign, which the grid does not find, so the page gives no answer.
3. Inside the domain: with h = 10⁻⁹ × max(1, |c|), a cut c is inside the domain of f when f(c) is a real number and |f(c ± h)| ≤ 10 |f(c ± 100h)| on both sides (f does not grow towards c).
4. Inflection points: the cuts c inside the domain where f″(c − h) and f″(c + h) have opposite signs (neither 0). The page lists their x values, in increasing order: "2", or "none".
5. Concavity: the cuts split the line into intervals (−∞, c₁), (c₁, c₂), …, (cₙ, ∞). In each, the sign of f″ is read at one point: the midpoint of a finite interval; c − max(1, |c|) for (−∞, c); c + max(1, |c|) for (c, ∞); 0 for (−∞, ∞) when there are no cuts. An interval whose point is not in the domain of f (f not a real number there) or where f″ is 0 is left out. Two neighbouring intervals with the same sign are joined when the cut between them is a zero of f″ (not a break) inside the domain (x⁴ is concave up on (-∞, ∞)). Intervals are not joined across a break of f″: x^(2/3)(x − 5) is concave up on (−1, 0) and (0, ∞), as f′ does not exist at 0, and x^(4/3) is concave up on (−∞, 0) and (0, ∞) (true on each part, though f″ > 0 on both sides of 0).

The page shows Inflection points at x =, Concave up on and Concave down on (open intervals separated by commas, with -∞ and ∞ written with a hyphen and the infinity sign, or none), and f″(x) = as the algebra writes it. Cut values are written exactly or as ≈ decimals, as below.

How the points are found

The zeros and breaks of a function g (a zero: g passes through or touches 0; a break: g jumps, goes to ±∞, or stops being a real number) are found between two ends −∞ and ∞ (for g = f″) (each end clipped to ±10⁶) in two ways at once.

By the algebra. The computer algebra system solves g(x) = 0. A solution is an exact candidate when it is a number strictly between the ends, is not a rounded number (a whole number past 2⁵³; a fraction p/q, p and q being the numbers that multiply its top and its bottom, such as 288557167/(342919925e), when q after removing its factors 2 and 5 is over 1,000,000, or is over 1 while |p| times it is over 10¹²; or a decimal with more than 12 significant digits), and g is 0 there up to rounding: |g(c)| is at most 10⁻¹² times the sum of the sizes of the terms of g at c (the parts of g joined by + and −). So x² + 10⁻²⁰ has no zero at 0, though the algebra gives one.

On a grid. g is worked out at 8,001 points x₀ to x₈₀₀₀, evenly spaced in asinh x (so they reach ±10⁶ when an end is infinite and bunch up near 0): xᵢ = sinh(u₀ + (u₁ − u₀)(i + 0.3183)/8001), with u₀ = asinh(−10⁶) and u₁ = asinh(10⁶). For each step from xᵢ to xᵢ₊₁:

• If g is a number at one end of the step and not a number at the other (as ln or √ of a negative number is not), the point where it stops being one is found by halving the step 80 times: a break. A value past the largest computer number, about 1.8 × 10³⁰⁸, still counts as a number with its sign (x e^(−x) and its derivatives for x below about −703), so an overflow is not a break. But when |g| is over 10³⁰⁰ at the end where g is a number (e^x − e^(2x) near x = 355, where e^(2x) overflows and the difference is not a number), the page cannot tell an overflow from the end of the domain and gives no answer (a value is too large).
• If g has opposite signs at the two ends, the sign change is found by halving 80 times. It is a zero if |g| there is at most 10⁻⁶ of the larger of |g(xᵢ)| and |g(xᵢ₊₁)|, and a break otherwise (g jumps or goes through ±∞). If g is past the largest computer number at either end, the page gives no answer (a value is too large).
• If g has one sign at xᵢ, xᵢ₊₁ and xᵢ₊₂ and turns at xᵢ₊₁ (a peak or a dip), the turning point r is found by halving (on the sign of g(x + h) − g(x − h), h = 10⁻⁷ × max(1, |x|)). If g(r) has the other sign, g dips through 0 and back between xᵢ and xᵢ₊₂: two zeros too close together for the grid. Unless the algebra gives at least two exact candidates strictly between xᵢ and xᵢ₊₂ (then they are the zeros), the page gives no answer (two points are too close together). If |g(r)| is at most 10⁻⁹ of the larger of |g(xᵢ)| and |g(xᵢ₊₂)|, g touches 0 there: that is a zero when an exact candidate lies within 10⁻⁶ × max(1, |candidate|) of r, and otherwise the page says a zero could not be confirmed and gives no answer. If |g(r)| is over 10⁶ times that size (or not a number), g peaks through ±∞: a break. A reason to give no answer that turns up on the way (a dip, a touch the algebra does not confirm) is given only after the whole grid, so a function with more than 30 points (sin x) gets "too many points to check".

A zero found on the grid within 10⁻⁶ × max(1, |c|) of an exact candidate c takes its exact value and text; otherwise, when it is within 10⁻⁹ × max(1, |x|) of a fraction p/q with q = 1, 2, 3, 4, 6, 8 or 12 (the first that fits) where g is 0 up to rounding (the rule above), it is shown as p/q; otherwise as ≈ and a decimal to 10 significant figures (rounded half up). A break is shown exactly as p/q when x is within 10⁻⁹ × max(1, |x|) of p/q for q = 1, 2, 3, 4, 6, 8 or 12 (the first that fits) and g is not a real number at p/q or at p/q ± 10⁻⁹ × max(1, |x|) (a pole, a jump, or an end of the domain); otherwise as ≈ and a decimal. An exact candidate that the grid did not find (two zeros inside one step) is added. A zero right next to a pole, inside the same grid step, is found only when the algebra gives it. The same point found twice counts once (a zero before a break). Two different points within 10⁻⁶ × max(1, |x|) of each other give no answer (two points are too close together), and so do two exact candidates that close: x²(x − 10⁻⁷)² has critical numbers at 0, 5 × 10⁻⁸ and 10⁻⁷, which the page does not merge into one. More than 30 points gives no answer (too many points to check).

What you can type

• A number has at most 15 digits in a row. A computer number keeps only about 16 digits, so a longer one (9007199254740993) would stand for a nearby number (9007199254740992), and the page asks for fewer digits instead. Write very large or very small numbers with a power of ten (1e-20).
• The function f uses the variable x. Numbers can have decimals (2.5) and powers of ten (1e-3).
• Operations: + − * / and ^ for powers. Brackets group. A number or bracket next to a letter multiplies: 2x, 3(x + 1), x sin(x).
• Constants: pi (or π) and e.
• Functions: sqrt, cbrt, ln (and log, the same natural logarithm), log10, exp, abs (or |x|), sin, cos, tan, sec, csc, cot, asin, acos, atan (arcsin, arccos, arctan also work), sinh, cosh, tanh and their inverses. Angles are in radians. sin x without brackets means sin(x); sin x^2 means sin(x²).

What gets no answer

• More than 30 cuts (sin(x)), or a possible zero of f″ that the algebra does not confirm.
• Two cuts too close together to tell apart (x⁴ − 10⁻¹²x², whose inflection points ±4.1 × 10⁻⁷ lie inside one grid step), a value past the largest computer number next to a cut, or an f″ that holds |u|.
• A step that fails its check, finds no formula, or takes over 3 seconds.

Worked examples by hand

f(x) = x³ − 6x² + 9x + 30 (OpenStax Calculus Volume 1, section 4.5, Example 4.19). f′(x) = 3x² − 12x + 9 and f″(x) = 6x − 12, which is 0 at x = 2. f″(0) = −12 < 0 and f″(3) = 6 > 0: f is concave down on (-∞, 2), concave up on (2, ∞), and the inflection point is at x = 2 (the point (2, 32), since f(2) = 8 − 24 + 18 + 30 = 32).

f(x) = x⁴ (the second derivative test for concavity, OpenStax Calculus Volume 1, section 4.5). f″(x) = 12x² is 0 at x = 0 and positive on both sides, so there is no inflection point and f is concave up on (-∞, ∞).