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Inflection points

Inflection points

An inflection point

An inflection point is a point in the domain of the function where the function changes its concavity – from concave up to concave down, or the other way round.

The sign of f′′\displaystyle f'' gives the concavity, so you look for inflection points like this:

  • Possible inflection points: points where f′′(x)=0\displaystyle {f''(x)}={0}, and points of the domain where f′′\displaystyle f'' is not defined.
  • The check: build a table of signs of f′′\displaystyle f''. A possible point is an inflection point only if the sign of f′′\displaystyle f'' changes there.
  • An inflection point is a point on the graph: work out f(x)\displaystyle f(x) there too.
Example: x3−3x2+2\displaystyle x^3-3x^2+2

The example below finds the inflection point of f(x)=x3−3x2+2\displaystyle {f(x)}={x^3-3x^2+2}:

  • f′(x)=3x2−6x\displaystyle {f'(x)}={3x^2-6x}, f′′(x)=6x−6\displaystyle {f''(x)}={6x-6}.
  • f′′(x)=0\displaystyle {f''(x)}={0} where x=1\displaystyle {x}={1}: a possible point.
  • f′′(x)<0\displaystyle f''(x)<0 where x<1\displaystyle x<1 and f′′(x)>0\displaystyle f''(x)>0 where x>1\displaystyle x>1: the sign changes, so there is an inflection.
  • f(1)=1−3+2=0\displaystyle {f(1)}={1-3+2}={0}: the inflection point is (1,0)\displaystyle (1,0).

The board also shows the tangent at the inflection point: f′(1)=−3\displaystyle {f'(1)}={-3}, and the tangent is y=−3x+3\displaystyle {y}={-3x+3}.

f(x)=x3−3x2+2\displaystyle f(x)=x^{3}-3x^{2}+2
y=−3x+3\displaystyle y=-3x+3
P=(1,0)\displaystyle P=(1,0)
  • Left of the inflection point the tangent is above the graph (the graph is concave down), and right of it the tangent is below the graph (the graph is concave up).
  • So at an inflection point the tangent crosses the graph.
A horizontal, a slanted and a vertical inflection

The board below shows three functions, each with an inflection point at the origin, each with a tangent of a different kind:

f(x)=x3\displaystyle f(x)=x^{3}
g(x)=x3−3x\displaystyle g(x)=x^{3}-3x
h(x)=x3\displaystyle h(x)=\sqrt[3]{x}
  • f(x)=x3\displaystyle {f(x)}={x^3}: f′(0)=0\displaystyle {f'(0)}={0}, the tangent is horizontal – a horizontal inflection.
  • g(x)=x3−3x\displaystyle {g(x)}={x^3-3x}: g′(0)=−3\displaystyle {g'(0)}={-3}, the tangent slants – a slanted inflection. This is the usual case: at most inflection points the first derivative is not 0.
  • h(x)=x3\displaystyle {h(x)}={\sqrt[3]{x}}: the tangent at the origin is vertical, and h′\displaystyle h' and h′′\displaystyle h'' are not defined there – a vertical inflection. It is a possible point of the second kind: h′′\displaystyle h'' is not defined there.

A horizontal inflection turns up already when you look for extremum points, since there f′(x)=0\displaystyle {f'(x)}={0}. A slanted inflection is found only with f′′\displaystyle f''.

f′′=0\displaystyle {f''}={0} with no inflection

A point where f′′(x)=0\displaystyle {f''(x)}={0} is not necessarily an inflection point. For example, f(x)=x4\displaystyle {f(x)}={x^4}:

  • f′′(x)=12x2\displaystyle {f''(x)}={12x^2}, so f′′(0)=0\displaystyle {f''(0)}={0}.
  • But f′′(x)>0\displaystyle f''(x)>0 on both sides of 0: the sign does not change.
  • The function is concave up everywhere, and (0,0)\displaystyle (0,0) is a minimum point, not an inflection point.
The inflection point of a cubic

Every function of degree three has exactly one inflection point. The proof below works out f′′\displaystyle f'' and shows that its sign changes exactly once.

  • f(x)=ax3+bx2+cx+d\displaystyle {f(x)}={ax^3+bx^2+cx+d}, with a≠0\displaystyle a\neq0.
  • f′(x)=3ax2+2bx+c\displaystyle {f'(x)}={3ax^2+2bx+c}, f′′(x)=6ax+2b\displaystyle {f''(x)}={6ax+2b}.
  • f′′\displaystyle f'' is a linear function with slope 6a≠0\displaystyle 6a\neq0: it is 0 at one point, x=−b3a\displaystyle {x}={-\frac{b}{3a}}, and its sign changes there.
  • So x=−b3a\displaystyle {x}={-\frac{b}{3a}} is the x\displaystyle x-coordinate of the only inflection point.
Exercise 1

The function is f(x)=x3−6x2+9x+1\displaystyle {f(x)}={x^3-6x^2+9x+1}.

a

Find f′′(x)\displaystyle f''(x).

f(x)=x3−6x2+9x+1\displaystyle f(x)=x^{3}-6x^{2}+9x+1

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

    b

    Solve the equation f′′(x)=0\displaystyle {f''(x)}={0}.

      c

      Mark the inflection point on the graph.

      f(x)=x3−6x2+9x+1\displaystyle f(x)=x^{3}-6x^{2}+9x+1

      Exercise 2

      The function is f(x)=x4−4x3\displaystyle {f(x)}={x^4-4x^3}.

      a

      Find f′′(x)\displaystyle f''(x).

      f(x)=x4−4x3\displaystyle f(x)=x^{4}-4x^{3}

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

        b

        Solve the equation f′′(x)=0\displaystyle {f''(x)}={0}.

          c

          Mark the inflection points on the graph.

          f(x)=x4−4x3\displaystyle f(x)=x^{4}-4x^{3}

          d

          f′(x)=4x3−12x2\displaystyle {f'(x)}={4x^3-12x^2}.
          At which of the inflection points is the tangent horizontal, and at which does it slant?

          Exercise 3

          The function is f(x)=x4+6x2\displaystyle {f(x)}={x^4+6x^2}.

          a

          Find f′′(x)\displaystyle f''(x).

          f(x)=x4+6x2\displaystyle f(x)=x^{4}+6x^{2}

            / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

            / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

            b

            Solve the equation f′′(x)=0\displaystyle {f''(x)}={0}.

              c

              Does f\displaystyle f have inflection points?
              What is the concavity of f\displaystyle f?

              Exercise 4

              The family of functions is f(x)=x3−ax2+3x\displaystyle {f(x)}={x^3-ax^2+3x}, where a\displaystyle a is a parameter.

              a

              Find f′′(x)\displaystyle f''(x), in terms of a\displaystyle a.
              Explain why every function of the family has one inflection point, at x=a3\displaystyle {x}={\frac{a}{3}}.

              f(x)=x3−ax2+3x\displaystyle f(x)=x^{3}-ax^{2}+3x

                / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                b

                For which value of a\displaystyle a is the inflection point at x=1\displaystyle {x}={1}?

                  c

                  For the a\displaystyle a you found, f′(x)=3x2−6x+3=3(x−1)2\displaystyle {f'(x)}={3x^2-6x+3}={3(x-1)^2}.
                  What kind of inflection is at x=1\displaystyle {x}={1}: horizontal, slanted or vertical?
                  Does the function have extremum points?

                  Exercise 5

                  Write a function that has an inflection point where the first derivative is positive.
                  Show that the point you chose is an inflection point, and work out the derivative there.