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Find and verify points of inflection using a change of concavity. Distinguish stationary and non-stationary inflections, reject false candidates and prove conditions for cubic and quartic curves.
Before you startSecond derivatives; sign charts; stationary points; function domains
01 / Check for a change of concavity
A point of inflection is a point on the curve where its concavity changes.
For the smooth functions in this lesson, find candidates by solving f″ = 0, then check the second derivative’s sign on both sides. A change from positive to negative, or negative to positive, confirms a change of concavity.
Substitute the candidate input into f to give the point’s coordinates. Then use f′ to decide whether it is stationary.
f′ = 3x²; f″ = 6x.
At x = −0.5, f″ = −3. At x = 0.5, f″ = 3. At zero: f′ = 0 and f″ = 0.
The second derivative changes from negative to positive: the origin is a stationary point of inflection.
Moving two sample points illustrates the signs; a proof uses the formula to establish the sign throughout an interval on each side. The gold tangent can be horizontal or sloping.
02 / Verify a stationary point of inflection
f′ = 3x²; f″ = 6x
Differentiate twice.
f″ = 0 gives x = 0
This is the candidate.
f″ < 0 for x < 0; f″ > 0 for x > 0
The concavity changes.
The point (0, 0) is a stationary inflection because f′(0) = 0
The first derivative is positive on both sides, so it is not a turning point.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Find and classify the point of inflection of y = (x − 2)³ + 5.
y″ = 6(x − 2).
The second derivative changes from negative to positive at x = 2. The point is (2, 5), and y′(2) = 0, so it is a stationary point of inflection.
03 / An inflection can have a nonzero gradient
f′ = 3x² + 1; f″ = 6x
The extra linear term changes the slope but not the second derivative.
f″ changes sign at x = 0
The origin is still an inflection.
f′(0) = 1
It is non-stationary, with tangent y = x.
Find the point of inflection and its tangent for y = x³ − 3x² + 2.
y″ = 6x − 6.
The sign changes at x = 1. The point is (1, 0), with gradient y′(1) = −3. The tangent is y = −3(x − 1). This is a non-stationary inflection.
04 / Reject a zero with no sign change
f′ = 4x³; f″ = 12x²
The origin is both stationary and a zero of f″.
f″ > 0 for every x ≠ 0
It is positive on both sides of zero.
There is no change of concavity, so no inflection at zero
The curve stays convex.
The point (0, 0) is a minimum
The first derivative changes from negative to positive.
Does y = x⁶ have an inflection at zero?
y″ = 30x⁴.
No. The second derivative is positive on both sides of zero. The origin is a minimum, not an inflection.
Does y = x⁵ have an inflection at zero?
y″ = 20x³.
Yes. The sign changes from negative to positive. Since y′(0) = 0, the origin is a stationary point of inflection.
05 / Check every candidate on a sign chart
y″ = 12x² − 12 = 12(x − 1)(x + 1)
The candidates are x = −1 and x = 1.
Signs: positive for x < −1, negative for −1 < x < 1, positive for x > 1
Both candidates give sign changes.
Inflection points: (−1, −5) and (1, −5)
Substitute into the original quartic.
Gradients: +8 and −8 respectively
Both are non-stationary.
Find all inflection points of y = x⁴ − 8x³ + 2x.
y″ = 12x(x − 4).
The second derivative changes sign at x = 0 and x = 4. The points are (0, 0) and (4, −248). The gradients are 2 and −126, so both are non-stationary.
06 / Use signs of positive factors
y′ = eˣ(x + 1); y″ = eˣ(x + 2)
Apply the product rule twice.
eˣ is always positive
The sign of y″ is the sign of x + 2.
Inflection at (−2, −2e⁻²)
The concavity changes from down to up.
y′(−2) = −e⁻² ≠ 0
This inflection is non-stationary.
Does y = ln x have any points of inflection on its real domain?
y″ = −1/x² and x > 0.
No. It is negative throughout the domain, so there is no change of concavity.
Find the interior point of inflection of y = sin x on 0 ≤ x ≤ 2π.
y″ = −sin x; inspect both sides of π.
The interior inflection is (π, 0), where the second derivative changes from negative to positive and y′ = −1. The domain endpoints are not interior two-sided inflections of this restricted curve.
07 / A gap is not a point of inflection
For y = 1/x, the second derivative is 2/x³. It is negative on x < 0 and positive on x > 0, but x = 0 is excluded and the graph is discontinuous there. There is no point of inflection at zero.
A sign change across a missing point is different from a concavity change at a point on the curve. Likewise, a domain endpoint does not have two neighbouring sides within the restricted curve.
Does y = 1/(x − 2) have an inflection at x = 2?
Its second derivative is 2/(x − 2)³.
No. Although the second derivative has opposite signs on the two domain intervals, the function is undefined at x = 2.
For the restricted curve y = x³ with x ≥ 0, is its endpoint (0, 0) an interior point of inflection?
There are no negative x-values in the stated domain.
No. The unrestricted cubic has an inflection there, but the restricted curve has no two-sided concavity change at its endpoint.
08 / Use the third derivative as a sufficient test
For a sufficiently smooth function:
f″(a) = 0 and f‴(a) ≠ 0 → an inflection at x = a
A nonzero third derivative means the second derivative crosses zero with a nonzero slope. The sign-change method remains the direct check. If f‴(a) = 0, the test says nothing by itself.
At zero, both x⁴ and x⁵ have f″ = 0 and f‴ = 0. Explain why their classifications differ.
Compare 12x² with 20x³.
For x⁴, f″ stays positive on both sides, so no inflection. For x⁵, f″ changes sign, so there is an inflection. A zero third derivative is not a rejection test.
09 / Prove that every genuine cubic has one inflection
f″ = 6ax + 2b
The second derivative is a nonconstant straight line.
Its unique zero is x = −b/(3a)
There is exactly one candidate.
The slope 6a is nonzero, so f″ changes sign there
The cubic has exactly one point of inflection.
f′ at this input is c − b²/(3a)
The inflection is stationary precisely when c = b²/(3a).
Find k so that y = 2x³ − 6x² + kx + 1 has a stationary point of inflection.
The inflection input is x = 1, independent of k.
y′(1) = k − 6, so k = 6. Then the stationary inflection is (1, 3).
10 / Count quartic inflections using a discriminant
f″ = 2(6ax² + 3bx + c)
The relevant quadratic has discriminant Δ = 9b² − 24ac.
If Δ > 0: two distinct real roots
The quadratic changes sign at each, so there are two inflections.
If Δ = 0: one repeated root
The second derivative touches zero without changing sign, so there is no inflection there.
If Δ < 0: no real roots
There are no inflections.
For y = x⁴ + ax², determine which values of a give points of inflection.
y″ = 12x² + 2a.
If a < 0, there are two inflections at x = ±√(−a/6), with y = −5a²/36. If a ≥ 0, there are none. At a = 0 the repeated zero is not a sign change.
Does y = x⁴ − 4x³ + 6x² have an inflection at x = 1?
y″ = 12(x − 1)².
No. The second derivative is nonnegative everywhere and positive on both sides of 1.
11 / Verify the sign change and report the point
Find and classify the inflection of y = 3x³ + 6x² + 7x.
y″ = 18x + 12.
x = −2/3, where y = −26/9 and y′ = 3. The second derivative changes from negative to positive, so (−2/3, −26/9) is a non-stationary point of inflection.
Section 1 of 11 · Check for a change of concavity