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The graph of a function and the graph of its derivative

The graph of a function and the graph of its derivative

What the graph of the derivative tells

The value of f′(x)\displaystyle f'(x) is the slope of the tangent to the graph of f\displaystyle f. So every feature of the graph of the derivative is a feature of the graph of the function:

  • The graph of the derivative above the x\displaystyle x-axis (f′>0\displaystyle f'>0) – the function rises.
  • The graph of the derivative below the x\displaystyle x-axis (f′<0\displaystyle f'<0) – the function falls.
  • The graph of the derivative crosses the x\displaystyle x-axis – the function has an extremum point: upwards (from − to +) – a minimum; downwards – a maximum.
  • The graph of the derivative touches the x\displaystyle x-axis without crossing it – the tangent is horizontal, but there is no extremum.

The derivative of a polynomial of degree n\displaystyle n is a polynomial of degree n−1\displaystyle n-1: the derivative of a function of degree three is a parabola, and the derivative of a parabola is a straight line.

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

The first board shows the graph of f(x)=x3−3x\displaystyle {f(x)}={x^3-3x}:

f(x)=x3−3x\displaystyle f(x)=x^{3}-3x

The second board shows the graph of its derivative, f′(x)=3x2−3\displaystyle {f'(x)}={3x^2-3}:

y=3x2−3\displaystyle y=3x^{2}-3
  • f\displaystyle f rises where x<−1\displaystyle x<-1 and where x>1\displaystyle x>1 – there the parabola is above the x\displaystyle x-axis.
  • f\displaystyle f falls where −1<x<1\displaystyle -1<x<1 – there the parabola is below the axis.
  • At the extremum points of f\displaystyle f, x=−1\displaystyle {x}={-1} and x=1\displaystyle {x}={1}, the parabola crosses the x\displaystyle x-axis.
  • At x=0\displaystyle {x}={0} the fall of f\displaystyle f is steepest, and there the parabola has its minimum, f′(0)=−3\displaystyle {f'(0)}={-3}.
From the graph of the derivative to the function

Even when only the graph of the derivative is known, it tells where the function rises and where it has an extremum. The board below shows the graph of the derivative f′(x)=(x+1)2(x−3)\displaystyle {f'(x)}={(x+1)^2(x-3)} of a function f\displaystyle f:

y=(x+1)2(x−3)\displaystyle y=(x+1)^{2}(x-3)
x\displaystyle xx<−1\displaystyle x<-1−1\displaystyle -1−1<x<3\displaystyle -1<x<33\displaystyle 3x>3\displaystyle x>3
f′(x)\displaystyle f'(x)−0−0+
f(x)\displaystyle f(x)fallsno extremumfallsminimumrises
  • At x=−1\displaystyle {x}={-1} the graph of the derivative touches the x\displaystyle x-axis – f\displaystyle f has a horizontal tangent there with no extremum.
  • At x=3\displaystyle {x}={3} it crosses the axis upwards – f\displaystyle f has a minimum there.
  • The graph of the derivative gives the x\displaystyle x-coordinate of the minimum, but not its y\displaystyle y-coordinate.
An antiderivative

An antiderivative of f\displaystyle f is a function F\displaystyle F whose derivative is f\displaystyle f: F′(x)=f(x)\displaystyle {F'(x)}={f(x)}. You find it by going backwards through the rule of differentiation – raise the power by 1 and divide by the new power:

(xn+1n+1)′\displaystyle \left(\frac{x^{n+1}}{n+1}\right)' =\displaystyle = xn\displaystyle x^n

For example, an antiderivative of f(x)=3x2\displaystyle {f(x)}={3x^2} is F(x)=x3\displaystyle {F(x)}={x^3}. But x3+5\displaystyle x^3+5 and x3−2\displaystyle x^3-2 are antiderivatives too: the derivative of a constant number is 0.

  • All the antiderivatives of f\displaystyle f have the same slopes at every x\displaystyle x, so their graphs are the same graph, shifted up or down.
  • So the antiderivative is not unique: the antiderivatives are all F(x)+C\displaystyle F(x)+C, where C\displaystyle C is any number.
A function from its derivative and a point

Of all the antiderivatives, only one passes through a given point. The point gives an equation for C\displaystyle C.

The example below finds f\displaystyle f, given that f′(x)=6x2−4x\displaystyle {f'(x)}={6x^2-4x} and that the graph of f\displaystyle f passes through the point (1,5)\displaystyle (1,5):

  • The antiderivatives: f(x)=2x3−2x2+C\displaystyle {f(x)}={2x^3-2x^2+C}.
  • f(1)=5\displaystyle {f(1)}={5}, so 2−2+C=5\displaystyle {2-2+C}={5}, and C=5\displaystyle {C}={5}.
  • f(x)=2x3−2x2+5\displaystyle {f(x)}={2x^3-2x^2+5}.
Exercise 1

The board below shows the graph of f(x)=x3−6x2\displaystyle {f(x)}={x^3-6x^2}.

a

Add the graph of f′\displaystyle f' to the board: write f′(x)\displaystyle f'(x) in a row of your own.

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

b

Where is the graph you wrote above the x\displaystyle x-axis?
What does the graph of f\displaystyle f do in that interval?

Exercise 2

The board below shows the graph of the derivative f′(x)=x2−2x−3\displaystyle {f'(x)}={x^2-2x-3} of a function f\displaystyle f.

a

Mark the points where the graph of the derivative meets the x\displaystyle x-axis.

y=x2−2x−3\displaystyle y=x^{2}-2x-3

b

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

    c

    At which x\displaystyle x does f\displaystyle f have a maximum, and at which a minimum?
    Explain by the graph of the derivative.

    maximum

      / – 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

      minimum

        / – 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

        Exercise 3

        The derivative of a function f\displaystyle f is f′(x)=x(x−2)2\displaystyle {f'(x)}={x(x-2)^2}.

        a

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

          b

          What are the extremum points of f\displaystyle f, and of what kind?
          What does f\displaystyle f have at x=2\displaystyle {x}={2}?

          Exercise 4

          The function is f(x)=4x3−6x+1\displaystyle {f(x)}={4x^3-6x+1}.

          a

          Write three different antiderivatives of f\displaystyle f.

          b

          Check one of them: write (F(x))′\displaystyle \left(F(x)\right)' with the function you chose, and differentiate.

            / – 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

            Exercise 5

            The derivative of a function f\displaystyle f is f′(x)=3x2−4\displaystyle {f'(x)}={3x^2-4}.
            The graph of f\displaystyle f passes through the point (1,2)\displaystyle (1,2).

            a

            Write all the antiderivatives of f′\displaystyle f'.

            b

            Substitute the point and find C\displaystyle C.

              c

              Write f(x)\displaystyle f(x) in a row of your own on the board, and check that the graph passes through the point.

              A=(1,2)\displaystyle A=(1,2)