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The graph of the derivative, and parameters

The graph of the derivative, and parameters

What the graph of the derivative tells

The graph of the derivative shows, at every x\displaystyle x, the slope of the tangent to the graph of the function. So it is read for the sign of f′\displaystyle f', not for the values of f\displaystyle f:

  • Where the graph of f′\displaystyle f' is above the x\displaystyle x-axis – the function f\displaystyle f increases.
  • Where the graph of f′\displaystyle f' is below the x\displaystyle x-axis – the function f\displaystyle f decreases.
  • The graph of f′\displaystyle f' crosses the x\displaystyle x-axis from above to below – f\displaystyle f has a maximum there; from below to above – a minimum.
  • The graph of f′\displaystyle f' touches the x\displaystyle x-axis without crossing it – the tangent to the graph of f\displaystyle f is horizontal, but there is no extremum there.
Example: reading the graph of the derivative

The drawing is the graph of the derivative f′(x)\displaystyle f'(x), not of f(x)\displaystyle f(x).
The graph of f′\displaystyle f' passes through the points (−2,0)\displaystyle (-2,0) and (1,0)\displaystyle (1,0).

y= ?\displaystyle y=\,?
(−2,0)\displaystyle (-2,0)
(1,0)\displaystyle (1,0)

The example below reads the properties of f\displaystyle f from the graph of f′\displaystyle f':

  • Where x<−2\displaystyle x<-2 the graph of f′\displaystyle f' is below the axis, so f\displaystyle f decreases.
  • Where x>−2\displaystyle x>-2 (except at x=1\displaystyle {x}={1}) the graph of f′\displaystyle f' is above the axis, so f\displaystyle f increases.
  • At x=−2\displaystyle {x}={-2} the graph of f′\displaystyle f' crosses the axis from below to above: f\displaystyle f has a minimum point there.
  • At x=1\displaystyle {x}={1} the graph of f′\displaystyle f' only touches the axis: the tangent to the graph of f\displaystyle f is horizontal there, but f\displaystyle f increases before it and after it – no extremum.
A sketch of the graph of the derivative

The example below sketches the graph of the derivative of f(x)=4x3−3x\displaystyle {f(x)}={4x^3-3x} from the properties of f\displaystyle f.

  • The function is odd: f(−x)=−4x3+3x=−f(x)\displaystyle {f(-x)}={-4x^3+3x}={-f(x)}.
  • (12,−1)\displaystyle \left(\frac{1}{2},-1\right) is a minimum point. The graph is symmetric about the origin, so (−12,1)\displaystyle \left(-\frac{1}{2},1\right) is a maximum point – with no more working.
  • f′(x)=0\displaystyle {f'(x)}={0} at the extremum points: x=−12\displaystyle {x}={-\frac{1}{2}} and x=12\displaystyle {x}={\frac{1}{2}}.
  • f′(x)>0\displaystyle f'(x)>0 where f\displaystyle f increases: x<−12\displaystyle x<-\frac{1}{2} and x>12\displaystyle x>\frac{1}{2}; f′(x)<0\displaystyle f'(x)<0 between them.
f(x)=4x3−3x\displaystyle f(x)=4x^{3}-3x
y=12x2−3\displaystyle y=12x^{2}-3

The graph of f′\displaystyle f' (the dashed one) is a parabola opening upwards, crossing the x\displaystyle x-axis at the two extremum points of f\displaystyle f. Indeed, f′(x)=12x2−3\displaystyle {f'(x)}={12x^2-3}.

A parameter in a tangent and in an extremum point

A parameter is a fixed number that is not known, written as a letter: a\displaystyle a, b\displaystyle b, k\displaystyle k. A fact about a tangent or an extremum point gives an equation, and its solution is the value of the parameter.

  • The slope of the tangent at x1\displaystyle x_1 is m\displaystyle m: the equation f′(x1)=m\displaystyle {f'(x_1)}={m}.
  • There is an extremum point at x1\displaystyle x_1: the equation f′(x1)=0\displaystyle {f'(x_1)}={0}.

In the example below f(x)=x3+ax2\displaystyle {f(x)}={x^3+ax^2} has an extremum point at x=2\displaystyle {x}={2}. We find a\displaystyle a:

f′(x)=3x2+2ax,f′(2)=12+4a=0,a=−3f'(x)=3x^2+2ax,\qquad f'(2)=12+4a=0,\qquad a=-3

In the second example the slope of the tangent to the graph of f(x)=ax2−4x\displaystyle {f(x)}={ax^2-4x} at x=1\displaystyle {x}={1} is 6:

f′(x)=2ax−4,f′(1)=2a−4=6,a=5f'(x)=2ax-4,\qquad f'(1)=2a-4=6,\qquad a=5
The extremum points after a shift and a reflection

A shift or a reflection moves the whole graph, and the extremum points with it – no differentiating again. f(x)=x3−3x\displaystyle {f(x)}={x^3-3x} has a maximum (−1,2)\displaystyle (-1,2) and a minimum (1,−2)\displaystyle (1,-2).

  • h(x)=f(x−2)\displaystyle {h(x)}={f(x-2)} – a shift of 2 to the right: a maximum (1,2)\displaystyle (1,2), a minimum (3,−2)\displaystyle (3,-2).
  • k(x)=−f(x)+3\displaystyle {k(x)}={-f(x)+3} – a reflection in the x\displaystyle x-axis and a shift of 3 up. The reflection turns a maximum into a minimum: (−1,2)\displaystyle (-1,2) goes to (−1,1)\displaystyle (-1,1), a minimum point, and (1,−2)\displaystyle (1,-2) goes to (1,5)\displaystyle (1,5), a maximum point.
Exercise 1

The drawing is the graph of the derivative f′(x)\displaystyle f'(x) of a function f(x)\displaystyle f(x).
The graph of f′\displaystyle f' meets the x\displaystyle x-axis at x=−2\displaystyle {x}={-2} and at x=3\displaystyle {x}={3}, and the y\displaystyle y-axis at the point (0,6)\displaystyle (0,6).

y= ?\displaystyle y=\,?
a

Where does f\displaystyle f increase, and where does it decrease?

b

What are the x\displaystyle x-coordinates of the extremum points of f\displaystyle f, and what is the kind of each?
Give your reasons.

c

Given: at x=1\displaystyle {x}={1} the value of f′\displaystyle f' is 6 too.
At which points is the slope of the tangent to the graph of f\displaystyle f equal to 6?
What can you say about the tangents at these points?

Exercise 2

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

a

Write its derivative function, y=f′(x)\displaystyle {y}={f'(x)}, on the graph board.

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

b

Where is the graph of f′\displaystyle f' below the x\displaystyle x-axis?
What happens to the function f\displaystyle f there?

Exercise 3

The function is f(x)=x3+ax+1\displaystyle {f(x)}={x^3+ax+1}.
The slope of the tangent to the graph at the point where x=1\displaystyle {x}={1} is 5.

a

Write an equation in a\displaystyle a and solve it.

    b

    Here is the graph with the value of a\displaystyle a you found.
    Write the equation of the tangent at the point (1,4)\displaystyle (1,4) on the graph board.

    f(x)=x3+2x+1\displaystyle f(x)=x^{3}+2x+1
    (1,4)\displaystyle (1,4)

    Exercise 4

    The function f(x)=x3+bx2\displaystyle {f(x)}={x^3+bx^2} has an extremum point at x=−2\displaystyle {x}={-2}.

    a

    Write an equation in b\displaystyle b and solve it.

      b

      Here is the graph with the value of b\displaystyle b you found.
      Mark its extremum points.

      f(x)=x3+3x2\displaystyle f(x)=x^{3}+3x^{2}

      c

      Say the kind of each extremum point.

      Exercise 5

      The function f(x)=ax3−3x\displaystyle {f(x)}={ax^3-3x} has a minimum point at x=1\displaystyle {x}={1}.

      a

      Write an equation in a\displaystyle a and solve it.

        b

        What are the coordinates of the minimum point?
        The function is odd: what is its other extremum point, and what is its kind?

        Exercise 6

        The function f(x)=−x3+3x2\displaystyle {f(x)}={-x^3+3x^2} has a minimum point (0,0)\displaystyle (0,0) and a maximum point (2,4)\displaystyle (2,4).

        a

        What are the extremum points of h(x)=f(x−1)\displaystyle {h(x)}={f(x-1)}, and what is the kind of each?

        b

        What are the extremum points of k(x)=−f(x)+2\displaystyle {k(x)}={-f(x)+2}, and what is the kind of each?

        c

        A new function is g(x)=f(x)+p\displaystyle {g(x)}={f(x)+p}.
        Find the p\displaystyle p for which the graph of g\displaystyle g touches the x\displaystyle x-axis at its maximum point.
        Write an equation in p\displaystyle p and solve it.