Calculus · Grade 12 and undergraduate

Concavity, Inflection Points and the Second Derivative Test

Quick answer

A graph is concave up where its slopes increase — it bends like a cup, with f″ > 0 — and concave down where its slopes decrease, with f″ < 0. An inflection point is where the concavity changes. At a critical point with f′(c) = 0, a positive second derivative means a local minimum and a negative one a local maximum; when f″(c) = 0 the test says nothing, and the first derivative test has to decide. In context, f″ tells whether a rate is speeding up or slowing down.

What you'll learn

  • Find where a graph is concave up and concave down
  • Locate inflection points
  • Classify critical points with the second derivative test, and know when it fails
  • Interpret the second derivative in context

How a graph bends

The first derivative tells which way a graph goes. The second derivative tells how it bends.

  • Concave up on an interval: the slopes increase, so f′f' is increasing and f′′>0f'' > 0. The graph bends like a cup and lies above its tangent lines.
  • Concave down: the slopes decrease, so f′′<0f'' < 0. The graph bends like a cap and lies below its tangent lines.

An inflection point is a point on the graph where the concavity changes. For a function with a second derivative, f′′f'' changes sign there.

Concave down, then concave up The graph of y = x cubed minus 3 x squared minus 9x plus 5. It bends downward to the left of x = 1 and upward to the right, changing at the inflection point (1, −6), between the local maximum at (−1, 10) and the local minimum at (3, −22). -2246-20-1010xy inflection (1, −6) max min
  • f(x) = x³ − 3x² − 9x + 5
Concave down, then concave up

The second derivative test

Second derivative test

Suppose f′(c)=0f'(c) = 0 and f′′f'' is continuous near cc.

  • If f′′(c)>0f''(c) > 0, then ff has a local minimum at cc.
  • If f′′(c)<0f''(c) < 0, then ff has a local maximum at cc.
  • If f′′(c)=0f''(c) = 0, the test is inconclusive.

Why the bend decides the extremum

At cc the tangent is horizontal. If f′′(c)>0f''(c) > 0, the slopes are increasing through cc: negative before it, 00 at it, positive after. That is exactly the −- to ++ pattern of the first derivative test, so cc is a minimum.

Picture it as a cup whose bottom sits at cc: a horizontal tangent in a curve that bends upward can only be at the bottom. The second derivative test is the first derivative test read off the bend. When f′′(c)=0f''(c) = 0 there is no bend to read, and anything can happen: x4x^4 has a minimum at 00, −x4-x^4 a maximum and x3x^3 neither, yet all three have f′′(0)=0f''(0) = 0.

The second derivative in context

The second derivative is the rate of change of a rate, and its sign says whether the first rate is speeding up or slowing down.

If P(t)P(t) is a population and P′(t)>0P'(t) > 0 with P′′(t)<0P''(t) < 0, the population is growing, but more and more slowly. The inflection point of a growth curve is the moment of fastest growth.

Worked examples

Common mistakes

Practice problems

  1. Where is f(x)=x3−6x2+4f(x) = x^3 - 6x^2 + 4 concave up? Concave down?

    Answer

    Up on (2,∞)(2, \infty); down on (−∞,2)(-\infty, 2)

    Full solution

    f′′(x)=6x−12f''(x) = 6x - 12, negative before 22 and positive after.

  2. Find the inflection point of f(x)=x3−6x2+4f(x) = x^3 - 6x^2 + 4.

    Answer

    (2,−12)(2, -12)

    Full solution

    f′′f'' changes sign at 22, and f(2)=8−24+4=−12f(2) = 8 - 24 + 4 = -12.

  3. Use the second derivative test on the critical points of f(x)=x3−6x2+4f(x) = x^3 - 6x^2 + 4.

    Answer

    Local max at x=0x = 0; local min at x=4x = 4

    Full solution

    f′(x)=3x2−12x=3x(x−4)f'(x) = 3x^2 - 12x = 3x(x - 4). f′′(0)=−12<0f''(0) = -12 < 0 and f′′(4)=12>0f''(4) = 12 > 0.

  4. Where is f(x)=e−x2f(x) = e^{-x^2} concave up?

    Answer

    For x<−12x < -\tfrac{1}{\sqrt{2}} and x>12x > \tfrac{1}{\sqrt{2}}

    Full solution

    f′(x)=−2xe−x2f'(x) = -2x e^{-x^2} and f′′(x)=(4x2−2)e−x2f''(x) = (4x^2 - 2)e^{-x^2}, positive when x2>12x^2 > \tfrac{1}{2}.

  5. Does f(x)=x4f(x) = x^4 have an inflection point?

    Answer

    No

    Full solution

    f′′(x)=12x2≥0f''(x) = 12x^2 \ge 0 everywhere, so the graph is never concave down. f′′(0)=0f''(0) = 0, but the concavity does not change.

  6. Find the inflection points of f(x)=sin⁡xf(x) = \sin x on (0,2π)(0, 2\pi).

    Answer

    (π,0)(\pi, 0)

    Full solution

    f′′(x)=−sin⁡xf''(x) = -\sin x, which is negative on (0,π)(0, \pi) and positive on (π,2π)(\pi, 2\pi).

  7. A company’s revenue R(t)R(t) satisfies R′(t)>0R'(t) > 0 and R′′(t)<0R''(t) < 0. Describe the revenue.

    Answer

    It is increasing, but more and more slowly.

    Full solution

    R′>0R' > 0 means increasing. R′′<0R'' < 0 means the rate of increase is itself decreasing.

  8. The graph of f′f' is increasing on (−∞,2)(-\infty, 2) and decreasing on (2,∞)(2, \infty). What does that say about ff at x=2x = 2?

    Answer

    ff has an inflection point at x=2x = 2.

    Full solution

    f′f' increasing means ff is concave up; decreasing means concave down. The change at 22 is an inflection point.

  9. Classify the critical points of f(x)=xln⁡xf(x) = x \ln x using the second derivative test.

    Answer

    A local minimum at x=1ex = \tfrac{1}{e}

    Full solution

    f′(x)=ln⁡x+1=0f'(x) = \ln x + 1 = 0 at x=1ex = \tfrac{1}{e}. f′′(x)=1x>0f''(x) = \tfrac{1}{x} > 0 there, so it is a local minimum, with value −1e-\tfrac{1}{e}.

  10. A student says f(x)=x4−4x3f(x) = x^4 - 4x^3 has a local extremum at 00 because f′(0)=0f'(0) = 0 and f′′(0)=0f''(0) = 0 means “flat”. What went wrong?

    Hint

    What does the second derivative test say when f′′(c)=0f''(c) = 0?

    Answer

    With f′′(0)=0f''(0) = 0 the test is inconclusive, and the first derivative test shows no extremum at 00.

    Full solution

    f′(x)=4x2(x−3)f'(x) = 4x^2(x - 3) is negative on both sides of 00, so ff decreases through 00.

    x=0x = 0 is a critical point with a horizontal tangent, but not a maximum or minimum.

Frequently asked questions

What does concave up mean?

The graph bends upward like a cup: its slopes increase, so f″ > 0, and it lies above its tangent lines.

What is an inflection point?

A point on the graph where the concavity changes, from up to down or down to up. It usually shows up as a sign change of f″.

What is the second derivative test?

If f′(c) = 0 and f″(c) > 0, f has a local minimum at c; if f″(c) < 0, a local maximum. If f″(c) = 0, the test is inconclusive.

Is every point with f″ = 0 an inflection point?

No. f(x) = x⁴ has f″(0) = 0 but is concave up on both sides, so there is no change and no inflection point.

What does the second derivative mean in context?

The rate of change of a rate. If sales are rising (S′ > 0) but S″ < 0, they are rising more and more slowly.

What to learn next

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