About the Inflection Point Calculator
This inflection point calculator finds where a curve changes from bending upward to bending downward, or the reverse. Type a function such as x^3 − 6x^2 + 9x + 1, x^4 − 6x^2 or x·e^(−x), choose the x-range to search, and it computes the second derivative f″(x), finds every place in the range where f″ changes sign, and returns each inflection point as an (x, y) pair.
It also splits the range into concavity intervals — concave up where f″ > 0 and concave down where f″ < 0 — and lists a test point in each, exactly like the sign chart you would draw for homework. Points where f″ = 0 without a sign change (as at x = 0 for x⁴) are flagged as not being inflection points, which is the most common mistake on exams.
It suits calculus students sketching curves, teachers building answer keys, and anyone analysing where growth starts to slow, such as the midpoint of a logistic curve. The second derivative is found symbolically and its roots are located numerically to about 10 significant figures; trig functions use radians.
How to use the inflection point calculator
- 1Type the function using x as the variable.
- 2Set the x-range to search (defaults to −10 to 10).
- 3Read the inflection points and the second derivative used.
- 4Check the concavity intervals and sign chart to see where the curve is concave up or down.
- 5Use the graph to confirm that f″ crosses zero at each inflection point.
Formula and method
The second derivative measures how the slope is changing. Where f″(x) > 0 the slope is increasing and the graph is concave up (cup-shaped); where f″(x) < 0 it is concave down. An inflection point is a point on the graph where the concavity switches, which happens where f″ crosses zero (or is undefined) and actually changes sign.
The calculator differentiates f twice symbolically, samples f″ at 2,001 evenly spaced points across your range, and refines every sign change by bisection. A zero of f″ without a sign change — such as x = 0 for f(x) = x⁴ — is reported but not counted. Sign changes at points where f itself is undefined (vertical asymptotes) are excluded. Very closely spaced roots narrower than the sampling step can be missed; narrow the range to resolve them.
- f″(x)
- Second derivative of f
- c
- x-coordinate of an inflection point
- (c, f(c))
- The inflection point on the graph
Worked examples
Cubic x³ − 6x² + 9x + 1
f′(x) = 3x² − 12x + 9 and f″(x) = 6x − 12, which is zero at x = 2 and changes from negative to positive there. f(2) = 8 − 24 + 18 + 1 = 3, so the inflection point is (2, 3).
Quartic x⁴ − 6x²
f″(x) = 12x² − 12 = 12(x − 1)(x + 1), which changes sign at x = ±1. Both points have y = 1 − 6 = −5. The curve is concave up outside (−1, 1) and concave down inside it.
x·e^(−x) on [0, 5]
f″(x) = (x − 2)e^(−x), which is negative before x = 2 and positive after. f(2) = 2e^(−2) ≈ 0.270671, so the inflection point is (2, 0.270671).
sin(x) from −1 to 7
f″(x) = −sin(x) changes sign at every multiple of π, so in [−1, 7] there are inflection points at x = 0, π ≈ 3.14159 and 2π ≈ 6.28319, all with y = 0.
x⁴ has no inflection point
f″(x) = 12x² is zero at x = 0 but positive on both sides, so the curve stays concave up — f″ = 0 alone is not enough for an inflection point.
Frequently asked questions
What is an inflection point?+
An inflection point is a point on a curve where it changes concavity — from concave up (bending like a cup) to concave down (like a cap) or vice versa. The function must be continuous there, and the second derivative changes sign.
How do you find inflection points?+
Compute the second derivative, solve f″(x) = 0 and find where f″ is undefined, then test the sign of f″ on each side of every candidate. Only candidates where the sign actually changes are inflection points; plug them into f to get the y-values.
Is every point where f″(x) = 0 an inflection point?+
No. For f(x) = x⁴, f″(0) = 0 but f″ is positive on both sides, so the graph is concave up throughout and there is no inflection point at 0. The sign change is what matters.
Can an inflection point occur where f″ is undefined?+
Yes, if f is continuous there. For f(x) = ∛x, f″ is undefined at 0 but changes from positive to negative, so (0, 0) is an inflection point. Where f itself is undefined, such as x = 0 for 1/x, there is no inflection point even if concavity changes.
What is the difference between a critical point and an inflection point?+
Critical points are where f′(x) = 0 or is undefined — candidates for maxima and minima. Inflection points are where f″ changes sign, marking a change in concavity. A point can be both, as at x = 0 for f(x) = x³.