© 2009  Rasmus ehf    and Jóhann Ísak

Functions 2

Lesson4

Investigating functions using derivatives



The derivative of a function is a measure of the gradient of the graph so we can therefore draw the following conclusions:

A function is increasing if the derivative is positive  (+) and decreasing if the derivative is negative (–).

This means that when the derivative changes sign from positive to negative or negative to positive there must be a turning point or vertex on the graph. These are called maximum and minimum points. They are not necessarily the greatest or smallest overall values of the function.

Looking at a table of signs of the derivative shows us where these stationary points occur.

Remember, a continuous function cannot change between negative and positive values without going through zero, therefore we look for stationary points by finding out when the derivative is 0.

Example 1

Find the derivative of  f(x) = x2, make a table of signs, and use this to draw the graph of f(x).

   If f(x) = x2 then f´(x) = 2x.

The stationary points occur when f´(x) = 2x = 0 , that is when x = 0. The following is a table showing the sign of f´(x)

 

                                 Y = x2

 The derivative is 0 when x = 0 and the gradient changes from – to + so  so this is a minimum point.

 

Example 2

Find the derivative of  f(x) = x3 – 3x2 + 4, make a table of signs for the derivative and use this to find the stationary points. Compare your results with the graph shown in your graphical calculator.

   If f(x) = x3 – 3x2 + 4 then f´(x) = 3x2 – 6x.

Find where the derivative is 0.

    3x2 – 6x = 0

   3x(x – 2) = 0

This equation has the solutions  x = 0 and x = 2 and the derivative changes sign in these points. We find the sign of  f´(x) = 3x2 – 6x.

We could have made the table without factorising first, simply by choosing values of x between the zeroes and find the sign of the derivative.

   f´(–1) = 3(–1)2 – 6(–1) = 3 + 6 = 9    (+)

   f´(1) = 3·12 – 6·1 = –3                       (–)

   f´(3) = 3·32 – 6·3 = 9                         (+)

This gives a simpler table than above but shows the same information.

From the table we can deduce that we have a maximum when x = 0. The gradient is 0 so the graph is horizontal and the gradient changes from + ( going up ) to  –  (going down).The derivative is also zero when  x = 2 and the gradient changes from – to + so here we have a minimum. We can find the maximum and minimum values of the function by putting these x values into the formula for the original function. 

   f(0) = 03 – 3·02 + 4 = 4

The function has a maximum value in the point  (0, 4).

   f(2) = 23 – 3·22 + 4 = 8 – 12 + 4 = 0

The function has a minimum value in the point (2, 0).

A graphical calculator shows the following graph.

 

Sometimes the derivative of a function is zero without the sign of the derivative changing as it goes through the zero point. In such cases there is no stationary point but what is called a point of inflexion.

Example 3

Let's examine the function f(x) = x3.

The derivative is f´(x) = 3x2 and is zero when x = 0.

The following is a table of signs for the derivative.

The point (0, 0) is a point of inflexion. The graph is increasing up to x = 0, is horizontal in x = 0 and then carries on increasing after 0.

 

Calculations using the derivative have many practical applications, particularly in finding maximum and minimum values. The following two examples  demonstrate this.

Example 4

We wish to make a cardboard box using a square piece of card with a side of 1 m. To do this we bend the corners as shown in the diagram.  How much do we need to cut out of the corners in order for  the box to have as big a volume as possible?

 

Call this x , which means the length of each side of the box will be 2x shorter than the card, that is 1 – 2x . The height of the box will also be x, and the volume V can be written as follows:

   V = height·length·breadth

       = x(1– 2x)(1 – 2x)

       = x(1 – 4x + 4x2)

       = x – 4x2 + 4x3

Differentiating this and finding when the derivative is 0 gives us:

   V´ = 1 – 8x + 12x2 = 0

This is a quadratic equation that can be solved by using a calculator or the quadratic formula.

It's obvious that we  can't have  x = ½,  as if we cut off ½ a metre there will be no box left. So this must give the minimum value of the volume. The maximum volume will be when x = .

   V() = (1 – 2·)(1 – 2·) = m3

The maximum volume of the box will be m3  when we cut  m from each corner.

 

Example 5

A rectangle is drawn as shown in the diagram. One side is formed by the line  y = 3 and one corner, P, lies on the graph of  f(x) = x2 . Find the coordinates of the point P in order for the rectangle to have the maximum possible area.

The sides of the rectangle are  x and  3 – y or 3 – x2 as the point P lies on the graph of f(x) = x2. The area is therefore

   A = length·breadth

      = x·(3 – x2) = 3x – x3

Diferentiating this and finding where the derivative is zero gives:

   A´ = 3 – 3x2 = 0

     3 = 3x2

     x = ±1

The rectangle is in the positive quadrant so we can't have  x = –1. The maximum area is when   x = 1 so we can put this value into the formula for the area.

   A = 3x – x3 = 3 – 1 = 2


Practise these methods then try  Quiz 4 on Functions 2.  
Remember to use the checklist to keep track of your work.