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Concavity and points of inflexion

  • In 8 of 20 recent exams
  • Paper 1 or 2
  • 1 to 4 marks
How it is set in the exam

Paper 1 (no calculator) and Paper 2 (calculator), 1–4 marks per part. Common (8 of 20 exam versions), usually straight after the stationary points in the same question, and often after "show that f′′(x)=…f''(x) = \ldots" (that part is Differentiate: choose the rule).

Spot it in the exam
Exam phrases to spot (5)

  • “find the x-coordinate of the point of inflexion”

  • “justify that it is a point of inflexion”

  • “find the values of xx for which the graph of ff is concave up”

  • “concave down”

  • “show that the graph of ff has no point of inflexion”

In disguise (4)
  • "the gradient of the graph is greatest (or least)": the maximum (or minimum) of f′f'
  • "the rate of change of NN is greatest", "NN is increasing most rapidly": the maximum of N′N'
  • "decreasing most rapidly": the minimum of f′f' (its most negative value)
  • "the gradient of the graph is increasing": f′′(x)>0f''(x) > 0, concave up
Looks like it, but isn't (5)
  • "has a point of inflexion at x=2x = 2; find aa": the inflexion is a given fact, so f′′(2)=0f''(2) = 0 is an equation for the constant, in Find unknown constants
  • the greatest value of NN itself: N′(t)=0N'(t) = 0, in Stationary points
  • a "nature" question whose answer is a stationary point of inflexion stays in Stationary points
  • inflexions read off a graph of f′f': Graphs of f, f′ and f″
  • the steepest point of a displacement–time graph (a motion): kinematics, Greatest or least velocity, speed, acceleration or displacement on an interval
Worked example

A cubic: find the inflexion, justify it, and the concavity

Paper 1 (no calculator)
Question

Let f(x)=x3−6x2+9x+1f(x) = x^3 - 6x^2 + 9x + 1.
(a) Find f′′(x)f''(x). [2]
(b) Find the coordinates of the point of inflexion of the graph of ff. [3]
(c) Justify that it is a point of inflexion. [2]
(d) Find the values of xx for which the graph of ff is concave down. [2]
(e) Find the least value of the gradient of the graph of ff. [2]

Try it yourself first, then compare.

Now try one like it

Write your working on paper or on screen. Paper Zero marks it step by step, like the worked example, and shows where each mark went. Free, with no account.

A handwritten page as Paper Zero marks it: ticks at the two lines that earned a mark, a cross on line 3 where +7 was copied as −7, and a short note beside that line.123456f′(x) = 4x − 124x − 12 = 0, so x = 3f(3) = 2(3)² − 12(3) − 7= 18 − 36 − 7 = −25minimum at (3, −25)M1 ✓A1 ✓A1 ✗11The +7 became −7: 18 − 36 + 7 = −11.
Steps

  1. Find f′′(x)f''(x)

    in a form whose sign is easy to read: factorised, with any always-positive factor (such as exe^x) set aside.

  2. Solve f′′(x)=0f''(x) = 0

    for the candidates. On : the zeros of f′′f'', or the turning points of the graph of y=f′(x)y = f'(x) (graph of the derivative).

  3. Decide which of three situations it is:

    • "Justify", "show that it is a point of inflexion", "is there a point of inflexion at …?", "find any points of inflexion": show that f′′f'' changes sign, with a sign table labelled f′′f'' or a value on each side. "f′′(c)=0f''(c) = 0" as the only reason scores .
    • "Show there is no point of inflexion": show that f′′f'' never changes sign, for example a positive factor times a quadratic with Δ<0\Delta < 0.
    • The question says there is one ("find the coordinates of the point of inflexion", "the point of inflexion PP"): f′′(c)=0f''(c) = 0 finds it, with no sign check needed, unless f′′(x)=0f''(x) = 0 has more than one root: then keep the roots where f′′f'' changes sign; if more than one is left, use the question's other information.

  4. The point:

    yy from ff. If the gradient there is asked for, f′(c)f'(c): zero for a stationary point of inflexion, non-zero otherwise.

  5. Concavity:

    concave up where f′′(x)>0f''(x) > 0 and concave down where f′′(x)<0f''(x) < 0, as intervals.

  6. Greatest or least rate:

    f′f' is greatest where f′′f'' changes from ++ to −-, and least where it changes from −- to ++. Pick the right one: "decreasing most rapidly" is the least (most negative) f′f'. On a given interval, compare with f′f' at the ends too.

Common mistakes

  • "f′′(c)=0f''(c) = 0, so it is a point of inflexion."

    Not a reason on its own: y=x4y = x^4 has f′′(0)=0f''(0) = 0 and no inflexion. Show the sign change.

  • Answering "the rate is greatest" with f′(x)=0f'(x) = 0.

    That finds where the value is greatest; the rate is greatest where f′′(x)=0f''(x) = 0.

  • Picking the wrong zero of f′′f''.

    For "decreasing most rapidly" you want the minimum of f′f'; another zero may be where it is increasing most rapidly.

More mistakes (2)

  • Concave up with the wrong derivative

    concave up is f′′(x)>0f''(x) > 0, not f′(x)>0f'(x) > 0.

  • "No point of inflexion" without the positive factor

    say why exe^x (or a square) can't be 00 before looking at the quadratic.

Keep going

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