Functions (cont'd)
ClementRadcliffe, Contributor

In this week's lesson, we will complete the review of functions. This is to be followed by the introduction to aspects of coordinate geometry. We will begin with the homework from last week.

• f and g are functions defined as follows:

f : x -- 3x - 5

g : x -- 1/2 x

(a) Calculate the value of f (-3)

(b) Write expressions for

(i) f ^-1(x)

(ii) g^-1 (x)

(c) Hence, or otherwise, write an expression for (gf)^-1

Solution

As f: x -- 3x - 5 ... f (x) = 3x- 5

(a) f(-3) = 3x - 3 - 5 = -9 -5 = -14.

... f (-3) = -14.

(b) (i) Since f (x) = 3x- 5 ... y = 3x- 5

... 3x = y + 5

... x = (y + 5)/3 Interchanging x for y

... y = (x + 5)/3

... f ^-1(x) = (x + 5)/3

(ii) Since g: x - 1/2x ... g(x) = 1/2x

... y = ?x

... x = 2y Interchanging x for y

... y = 2x

... g ^-1 (x) = 2x

(c) (gf) ^-1(x) = f ^-1 g ^-1 (x)

This relationship may be given f (x) and g (x)

Since g ^-1 (x) = 2x, ... f ^-1 g ^-1 (x) = f ^-1 (2x)

N.B. Since f ^-1(x) = (x + 5)/3,

... f ^-1 (2x) = (2x + 5)/3

N.B. 2x replaces x in f ^-1(x)

... (gf) ^-1(x) = (2x + 5)/3

We will now begin to review co-ordinate geometry by considering straight lines on the Cartesian plane with respect to the following:

Gradient	Intercept
Mid-point	Length of line
Equation of line

Again, let me remind you of the importance of the theory of graphs as it is very important to this topic.

The Cartesian plane consists of the perpendicular x and y axes.

Reminders

• The axes must be properly labelled.
• Appropriate scales should be accurately used.
• The coordinates of a point are always expressed in the form: (x , y).
• Three points are required to draw a straight line. A ruler MUST ALWAYS be used to join the points.

GRADIENT

The gradient of a line is a measure of its slope. The value is denoted by m and is defined as:

m = Increase in the y coordinates/Increase in the x coordinates

Given two points represented by A (x1 , y1), and B (x2, y2) then the formula is:

m = y2 - y1/ x2 - x1

Example

Find the gradient m of the line joining the points A(2, 5), B(I, 2).

Since m = y2 - y1/x2 - x1, substituting

m = 2 - 5/1- 2 = -3/-1 = 3.

Answer = 3.

MID-POINT

This point is denoted by M and from the diagram, the coordinates of the mid-point are:

M = (x2 + x1)/2, (y2 + y1)/2

Example:

Find the coordinates of M, mid-point of A(2, 5) and B(1, 2).

Substituting into the formula above:

... M = (1 + 2)/2 , (2 + 5)/2

Answer: 3/2 , 7/2

Given the points A(x1, y1) and B(x2, y2), then finding the gradient and mid-point involves substituting into the appropriate formulae. Please note each formula well and always ensure that you use the correct one.

The following will illustrate this:

Given the points A(2, -3) and B(-4, 1), find:

(i) the gradient of AB

(ii) the mid-point of AB

Solution

(i) Gradient of AB = m = y2 - y1/ x2 - x1

Substituting the coordinates

... m = (1 - -3)/-4 - 2 = (1 + 3)/ - 6= -4/6.

... m = - 2/3

(ii) The mid-point of AB = M = (x2 + x1)/2, (y2 + y1)/2 Substituting

... M = (2 - 4)/2 , (-3 + 1)/2 = - 2/2 , - 2/2

... M = (-1 , -1)

Homework

Given the points X(-5, 3) and Y(1, 1), find the values of:

(a) Gradient, m

(b) the coordinates of the mid-point, M.

John Ricketts (right) and a schoolmate sing a tune titled: 'I am scared' during Peace Day activities at Penwood High School in St Andrew on Tuesday, March 4. -Anthony Minott/Freelance Photographer

Clement Radcliffe is the principal of Glenmuir High School in May Pen.