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🌌 Calculus Kingdom · Calculus

Concavity & Curve Sketching

Use the second derivative to read the bend of a graph: concave up or down, inflection points, the second derivative test, and the shape that the signs of f′ and f″ together force — then assemble the features of a curve before drawing it.

In short

  • f' tells you the direction of a curve and f'' tells you its bend. Concave up means f'' > 0 and the curve holds water; concave down means f'' < 0.
  • An inflection point needs f'' to be zero AND to change sign there. f(x) = x4 has f''(0) = 0 and no inflection point, which is why the sign check is not optional.
  • The second derivative test at a stationary point c: f''(c) > 0 is a local minimum, f''(c) < 0 is a local maximum, and f''(c) = 0 gives no verdict at all — go back to the sign of f' on each side.
  • A sketch is a checklist, not a guess: domain, intercepts, asymptotes, the sign chart for f', the sign chart for f'', then join the marked points.

Bend is a different question from direction

A road can be climbing and still be levelling off. It can be falling and still be falling faster and faster. Which way a curve is going and which way it bends are two separate facts, and calculus answers them with two separate derivatives.

  • f'(x) answers: is the curve going up or down as x increases?
  • f''(x) answers: which way does the curve bend?

When f''(x) > 0 the curve is concave up. It bends like the inside of a bowl, or a cup that would hold water. When f''(x) < 0 it is concave down: a dome, an upturned cup, water runs off it.

There is a second way to say the same thing that is worth carrying around. f'' is the derivative of f', so f'' > 0 means f' is increasing — the slope itself is getting bigger. On a concave-up stretch the slopes run -5, -2, 0, 3, 7: still rising, whether the curve is going up or down. On a concave-down stretch the slopes shrink.

That is why a graph of f' is enough to answer a question about the concavity of f. You do not look at whether f' is above or below the axis (that is direction); you look at whether the f' graph is rising or falling.

Inflection points: where the bend reverses

An inflection point is a point where the concavity changes — concave up on one side, concave down on the other. It is the moment a curve stops opening one way and starts opening the other.

Since concavity is the sign of f'', an inflection point needs the sign of f'' to change. That gives a two-step method, and the second step is the one people skip:

1. Solve f''(x) = 0 (or find where f'' has no value). 2. Check that f'' actually changes sign there.

Step 2 is not a formality. Take f(x) = x4. Then f''(x) = 12x2, which is zero at x = 0 — but 12x2 is positive on both sides, so the curve is concave up throughout and there is no inflection point at all. The graph flattens at the origin and carries on bending the same way.

Compare f(x) = x3, where f''(x) = 6x. That one really does go from negative to positive at 0, so the origin is an inflection point.

An inflection point is a point, so it has two coordinates. Solving f''(x) = 0 gives you only the x. The height has to come from substituting that x back into the original f — never into f' or f''.

Worth knowing: for any cubic ax3 + bx2 + cx + d, f'' = 6ax + 2b is a straight line with non-zero slope, so it always crosses the axis exactly once. Every cubic has exactly one inflection point. A quartic can have two, one or none.

The second derivative test

At a stationary point — a value c where f'(c) = 0 — the concavity settles what kind of point it is, and it does so with one substitution instead of a whole sign chart.

The second derivative test. Suppose f'(c) = 0. Then:

  • f''(c) > 0: the curve is concave up at c, so c sits at the bottom of a bowl. Local minimum.
  • f''(c) < 0: the curve is concave down at c, so c sits at the top of a dome. Local maximum.
  • f''(c) = 0: the test tells you nothing. Not "neither", not "an inflection point" — nothing.

The picture makes the first two easy to remember and hard to get backwards. A bowl holds water and its lowest point is the minimum; a dome sheds water and its highest point is the maximum. Positive means bowl, so positive means minimum.

The third case is the honest one. At c the test simply has no verdict, and all three outcomes are still possible:

  • f(x) = x4 at c = 0: f'(0) = 0 and f''(0) = 0, and 0 is a genuine local minimum.
  • f(x) = -x4 at c = 0: same two zeros, and 0 is a local maximum.
  • f(x) = x3 at c = 0: same two zeros again, and 0 is neither — the curve keeps climbing straight through.

When the test is silent, fall back on the first derivative test: check the sign of f' just to the left and just to the right of c. Positive-then-negative is a maximum, negative-then-positive is a minimum, and the same sign on both sides is neither.

Four shapes, and only four

Between them, the two signs give four possible local shapes. Learning them as pictures is faster than re-deriving them each time.

f' > 0, f'' > 0 rising, bending upwards climbing and getting steeper f' > 0, f'' < 0 rising, bending downwards climbing but levelling off f' < 0, f'' > 0 falling, bending upwards falling but flattening out f' < 0, f'' < 0 falling, bending downwards falling and getting steeper

Read the middle row in words: a population that is still growing, but growing more slowly every year. Read the third row: a cooling cup of tea, still losing heat but losing it more slowly. The signs of the two derivatives are exactly the mathematics of those sentences.

The two questions never answer each other. A curve that is concave up can be going down. A curve that is going up can be concave down. If you find yourself concluding "f'' < 0, so the curve is decreasing", you have handed f'' a job that belongs to f'.

A checklist for sketching a curve

A sketch is not guesswork. Collect the features first, mark them on the axes, and only then join them up. Everything below is something you can compute.

1. Domain. Where does the rule have a value at all? A denominator of zero or a negative under a square root cuts holes in the picture. 2. Intercepts. f(0) gives the y-intercept. Solving f(x) = 0 gives the x-intercepts, when they are findable. 3. Asymptotes, for a rational function. - Vertical: where the denominator is zero and the numerator is not. If both are zero, you have a hole rather than an asymptote. - Horizontal or slant: compare degrees. Bottom outranks top, the graph settles on y = 0. Equal degrees, it settles on the ratio of the leading coefficients. Top exactly one degree higher, divide out and the polynomial part is a slant asymptote. 4. f' = 0: the critical numbers, and a sign chart for f' giving the intervals where f rises and falls. 5. f'' = 0: the candidate inflection points, and a sign chart for f'' giving the intervals of concavity. 6. Classify each stationary point with the second derivative test, falling back on the first derivative test if f'' comes out zero. 7. Draw: dashed asymptotes first, then the marked points, then a curve joining them that obeys every sign you found.

Two habits make the sketch reliable. Test a value inside each interval rather than trusting a shape you expect, and check the far ends: what happens as x grows large positive, and as it grows large negative?

Worked examples

Example 1

Sketch f(x) = x3 - 3x2 - 9x + 5 by collecting its features: intervals of increase and decrease, local extrema, concavity and the inflection point.

  1. y-intercept: f(0) = 5, so the curve passes through (0, 5). The domain is every real number, and a cubic has no asymptotes.
  2. Differentiate: f'(x) = 3x2 - 6x - 9 = 3(x + 1)(x - 3). So the critical numbers are x = -1 and x = 3.
  3. Sign of f': test x = -2 gives 3(-1)(-5) = 15 > 0; test x = 0 gives 3(1)(-3) = -9 < 0; test x = 4 gives 3(5)(1) = 15 > 0. So f is increasing when x < -1, decreasing when -1 < x < 3, and increasing when x > 3.
  4. Differentiate again: f''(x) = 6x - 6 = 6(x - 1).
  5. Second derivative test at x = -1: f''(-1) = -12, which is negative, so the curve is concave down there and x = -1 is a local maximum. Its height is f(-1) = -1 - 3 + 9 + 5 = 10.
  6. Second derivative test at x = 3: f''(3) = 12, which is positive, so x = 3 is a local minimum. Its height is f(3) = 27 - 27 - 27 + 5 = -22.
  7. Concavity: f''(x) = 0 at x = 1, and 6(x - 1) is negative to the left and positive to the right, so f is concave down when x < 1 and concave up when x > 1. The sign genuinely changes, so x = 1 is an inflection point, at height f(1) = 1 - 3 - 9 + 5 = -6.
  8. Sketch: come up from the lower left, peak at (-1, 10), fall through the inflection point (1, -6) where the bend flips from dome to bowl, bottom out at (3, -22), and climb away to the upper right.

Example 2

Collect the features needed to sketch the rational function f(x) = (x2 + 4)/x.

  1. Domain: the denominator is zero at x = 0, so the graph exists for every x except 0.
  2. Rewrite it as a sum of powers, which makes everything after this easier: f(x) = x + 4/x = x + 4x-1.
  3. Vertical asymptote: the denominator is zero at x = 0 and the numerator there is 4, which is not zero, so x = 0 is a vertical asymptote rather than a hole.
  4. Slant asymptote: the top is degree 2 and the bottom degree 1, one higher, so there is a slant asymptote. From f(x) = x + 4/x, the fraction 4/x shrinks towards nothing as x grows, leaving y = x.
  5. Intercepts: there are none. x2 + 4 is never zero, and x = 0 is not in the domain.
  6. Critical numbers: f'(x) = 1 - 4x-2 = 1 - 4/x2. Setting that to zero gives x2 = 4, so x = 2 and x = -2.
  7. Classify them: f''(x) = 8x-3 = 8/x3. Then f''(2) = 1, which is positive, so x = 2 is a local minimum, at height f(2) = 2 + 2 = 4. And f''(-2) = -1, which is negative, so x = -2 is a local maximum, at height f(-2) = -4.
  8. Concavity: 8/x3 is positive when x > 0 and negative when x < 0, so the right-hand branch is concave up throughout and the left-hand branch is concave down throughout. There is no inflection point, because the only sign change happens at x = 0, where the function has no value.
  9. Sketch: two branches. For x > 0 the curve comes down the asymptote, turns at the minimum (2, 4) and rises alongside y = x. For x < 0 it is the same picture rotated through half a turn, with a maximum at (-2, -4).

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

f(x) = 3x3 - 3x2 + 2x - 2 Find f''(x), the second derivative. Give your answer as an expression in x.

Answer: 18x - 6

  1. f'(x) = 9x2 - 6x + 2
  2. f''(x) = 18x - 6

Problem 2

Difficulty 3 of 5

f(x) = 3x3 - 54x2 - x + 1 On which x-values is f concave up, and on which is it concave down?

  1. concave up when x > 6, concave down when x < 6
  2. concave up when x > -6, concave down when x < -6
  3. concave up when x < 6, concave down when x > 6
  4. concave up for every value of x

Answer: A. concave up when x > 6, concave down when x < 6

  1. f'(x) = 9x2 - 108x - 1
  2. f''(x) = 18x - 108 = 18(x - 6)
  3. f''(x) = 0 only at x = 6.
  4. The factor 18 is positive, so f'' is negative to the left of 6 and positive to the right.
  5. So f is concave up when x > 6, concave down when x < 6.

Problem 3

Difficulty 4 of 5

f(x) = 2x3 + 18x2 + 5x - 5 Find the inflection point of this curve. Give your answer as the pair (x, y).

Answer: (-3, 88)

  1. f'(x) = 6x2 + 36x + 5
  2. f''(x) = 12x + 36
  3. 12x + 36 = 0 gives x = -3.
  4. f(-3) = 88
  5. Inflection point: (-3, 88)

Common mistakes

  • Reading concavity off the first derivative — saying "f' < 0, so the curve is concave down". The sign of f' only says the curve is falling; it could be falling like a bowl or like a dome. Bend always comes from f''.
  • Calling every solution of f''(x) = 0 an inflection point. For f(x) = x4 the second derivative 12x2 is zero at x = 0 but positive on both sides, so the bend never reverses. Always test the sign of f'' on each side of the candidate.
  • Getting the second derivative test backwards: answering "maximum" when f''(c) is positive. A positive second derivative bends the curve upwards into a bowl, and the bottom of a bowl is a minimum.
  • Answering "neither" or "it is an inflection point" when the second derivative test comes out zero. Zero means the test has no verdict — the point may still be a maximum, a minimum or neither, and only the sign of f' either side will say which.
  • Giving only the x-value when an inflection point is asked for. Solving f''(x) = 0 gives the x; the height has to come from substituting that x back into the original f, not into f' or f''.

What you should be able to do

  • Find the second derivative and the intervals of concavity.
  • Find inflection points and classify stationary points with the second derivative test.
  • Match the shape of a graph to the signs of the first and second derivatives.
  • Collect the features of a curve — intercepts, asymptotes, extrema, concavity — for a sketch.

Where this fits in the curriculum

Common Core

  • FUN-4.A

    AP Calculus AB, Unit 5 — Justify conclusions about the behaviour of a function based on the behaviour of its derivatives: concavity, inflection points, the second derivative test, sketching.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

  • FUN-3.F

    AP Calculus AB, Unit 3 — Determine higher-order derivatives of a function.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

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