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Where a function rises and falls, and extremum points

Where a function rises and falls, and extremum points

The sign of the derivative

The derivative at a point is the slope of the tangent there, so its sign tells which way the graph goes:

  • f′(x)>0\displaystyle f'(x)>0 – the tangent goes up, and the function rises;
  • f′(x)<0\displaystyle f'(x)<0 – the tangent goes down, and the function falls;
  • f′(x)=0\displaystyle {f'(x)}={0} – the tangent is horizontal.
f′(x)>0\displaystyle f'(x)>0

Where the function rises is the solution of the inequality f′(x)>0\displaystyle f'(x)>0, and where it falls – of f′(x)<0\displaystyle f'(x)<0.

A point that may be an extremum

At a maximum point the function stops rising and starts falling, and at a minimum point – the other way round. At both the tangent is horizontal, so every extremum point is a solution of the equation:

f′(x)\displaystyle f'(x) =\displaystyle = 0\displaystyle 0
  • A solution of f′(x)=0\displaystyle {f'(x)}={0} is a point that may be an extremum.
  • A table of signs decides: check the sign of f′\displaystyle f' in each interval between these points.
  • f′\displaystyle f' goes from + to −: a maximum point. From − to +: a minimum point. The same sign on both sides: no extremum.
Example: f(x)=x3−6x2+9x\displaystyle {f(x)}={x^3-6x^2+9x}

f(x)=x3−6x2+9x=x(x−3)2\displaystyle {f(x)}={x^3-6x^2+9x}={x(x-3)^2} is the function from the first lesson. There we read its extremum points off the graph; the example below finds them by working them out.

f′(x)=3x2−12x+9=3(x−1)(x−3)f'(x)=3x^2-12x+9=3(x-1)(x-3)

The derivative is 0 at x=1\displaystyle {x}={1} and at x=3\displaystyle {x}={3}. We substitute one number from each interval in the derivative – say f′(0)=9\displaystyle {f'(0)}={9}, f′(2)=−3\displaystyle {f'(2)}={-3}, f′(4)=9\displaystyle {f'(4)}={9} – and get the table:

x\displaystyle xx<1\displaystyle x<11\displaystyle 11<x<3\displaystyle 1<x<33\displaystyle 3x>3\displaystyle x>3
f′(x)\displaystyle f'(x)+0−0+
f(x)\displaystyle f(x)risesmaximumfallsminimumrises
f(x)=x3−6x2+9x\displaystyle f(x)=x^{3}-6x^{2}+9x
A=(1,4)\displaystyle A=(1,4)
B=(3,0)\displaystyle B=(3,0)
  • The function rises where x<1\displaystyle x<1 and where x>3\displaystyle x>3, and falls where 1<x<3\displaystyle 1<x<3.
  • f(1)=1−6+9=4\displaystyle {f(1)}={1-6+9}={4}: a maximum point (1,4)\displaystyle (1,4).
  • f(3)=27−54+27=0\displaystyle {f(3)}={27-54+27}={0}: a minimum point (3,0)\displaystyle (3,0).
A derivative of 0 with no extremum

Not every point where f′(x)=0\displaystyle {f'(x)}={0} is an extremum point. The example below checks f(x)=x3−3x2+3x\displaystyle {f(x)}={x^3-3x^2+3x}:

  • f′(x)=3x2−6x+3=3(x−1)2\displaystyle {f'(x)}={3x^2-6x+3}={3(x-1)^2}, which is 0 only at x=1\displaystyle {x}={1}.
  • On both sides of 1 the derivative is positive: (x−1)2>0\displaystyle (x-1)^2>0.
  • So the function rises before 1 and after it too. At the point (1,1)\displaystyle (1,1) the tangent is horizontal, but there is no extremum.
f(x)=x3−3x2+3x\displaystyle f(x)=x^{3}-3x^{2}+3x
A=(1,1)\displaystyle A=(1,1)
Local and absolute extremum, and an extremum at the end

An extremum point is local: its value is the greatest (or the least) only close around it. The greatest value in the whole domain is the absolute maximum, and the least – the absolute minimum.

  • x3−6x2+9x\displaystyle x^3-6x^2+9x, defined for every x\displaystyle x, has no absolute maximum: as x\displaystyle x grows, f(x)\displaystyle f(x) grows without end.
  • When the domain is closed, say 0≤x≤5\displaystyle 0\leq x\leq5, there is also an extremum at the end at each end of the interval.

The example below finds the extremum points of f(x)=x3−6x2+9x\displaystyle {f(x)}={x^3-6x^2+9x} in the interval 0≤x≤5\displaystyle 0\leq x\leq5. We work out the values at the ends and at the points that may be extrema:

  • f(0)=0\displaystyle {f(0)}={0}: from 0 to 1 the function rises, so (0,0)\displaystyle (0,0) is a minimum point at the end.
  • f(1)=4\displaystyle {f(1)}={4}: a local maximum point.
  • f(3)=0\displaystyle {f(3)}={0}: a local minimum point.
  • f(5)=125−150+45=20\displaystyle {f(5)}={125-150+45}={20}: from 3 to 5 the function rises, so (5,20)\displaystyle (5,20) is a maximum point at the end.

The absolute maximum in the interval is 20, at x=5\displaystyle {x}={5}. The absolute minimum is 0, reached twice: at x=0\displaystyle {x}={0} and at x=3\displaystyle {x}={3}.

Exercise 1

The function is f(x)=x3−3x2−9x+5\displaystyle {f(x)}={x^3-3x^2-9x+5}.

a

Find f′(x)\displaystyle f'(x).

f(x)=x3−3x2−9x+5\displaystyle f(x)=x^{3}-3x^{2}-9x+5

    / – 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

      Solve the inequality f′(x)>0\displaystyle f'(x)>0: where does f\displaystyle f rise?

        d

        Mark the extremum points of f\displaystyle f on the graph.

        f(x)=x3−3x2−9x+5\displaystyle f(x)=x^{3}-3x^{2}-9x+5

        e

        Write where f\displaystyle f falls.
        Write the extremum points and the kind of each.

        Exercise 2

        The function is f(x)=−x3+12x\displaystyle {f(x)}={-x^3+12x}.

        a

        Write an inequality and solve it: where does f\displaystyle f rise?

          b

          Mark the extremum points of f\displaystyle f on the graph.

          f(x)=−x3+12x\displaystyle f(x)=-x^{3}+12x

          c

          Decide the kind of each extremum point.
          Explain by where the function rises, as you found.

          Exercise 3

          The function is f(x)=x4−8x2+3\displaystyle {f(x)}={x^4-8x^2+3}.

          a

          Write the equation f′(x)=0\displaystyle {f'(x)}={0} and solve it.

            b

            Build a table of signs of f′(x)\displaystyle f'(x).
            Write where the function rises and falls, and its extremum points and their kinds.

            c

            Mark the extremum points on the graph.

            f(x)=x4−8x2+3\displaystyle f(x)=x^{4}-8x^{2}+3

            Exercise 4

            The function is f(x)=x3+3x\displaystyle {f(x)}={x^3+3x}.

            a

            Solve the inequality f′(x)>0\displaystyle f'(x)>0.

              b

              Does f\displaystyle f have extremum points?
              Is there anywhere it falls?
              Explain.

              Exercise 5

              The function f(x)=x3−3x2\displaystyle {f(x)}={x^3-3x^2} is defined in the interval −1≤x≤4\displaystyle -1\leq x\leq4.

              a

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

                b

                Work out the values of the function at the ends of the interval and at the points you found.

                f(−1)\displaystyle f(-1)

                f(x)=x3−3x2\displaystyle f(x)=x^{3}-3x^{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

                  left extremum

                  f(x)=x3−3x2\displaystyle f(x)=x^{3}-3x^{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

                    right extremum

                    f(x)=x3−3x2\displaystyle f(x)=x^{3}-3x^{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

                      f(4)\displaystyle f(4)

                      f(x)=x3−3x2\displaystyle f(x)=x^{3}-3x^{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

                        c

                        Write all the extremum points of f\displaystyle f in the interval, at the ends too, and the kind of each.
                        What is the absolute maximum, and what is the absolute minimum?