Reviewing functions
Clement Radcliffe, 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 last week's homework.

• f and g are functions defined as follows

f: x ---> (x + 1)/2

g : x ---> 2x + 7

(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

(a) As f: x ---> (x + 1)/2

... f(x) = (x + 1)/2

... f(-3) = (-3 + 1)/2 = -1

... f(-3) = -1

(b) (i) Since f: x ---> (x + 1)/2

... y = (x + 1)/2

... 2y = x + 1
... x = 2y - 1 Interchanging x for y
... y = 2x - 1
... f^-1(x) = 2x - 1

(ii) Since g : x ---> 2x + 7

... g(x) = 2x + 7
... y = 2x + 7
... 2x = y - 7
... x = (y - 7)/2 Interchanging x for y
... y = (x - 7)/2

... g^-1(x) = (x - 7)/2

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

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

Since g^-1(x) = (x - 7)/2

... f^-1g^-1(x) = f^-1 ( (x-7)/2 )

NB Since f^-1(x) = 2x - 1

... f^-1 ( (x - 7)/2 ) = 2 x ( (x - 7)/2 ) -1

= x - 7 - 1 = x - 8

... f^-1 ( (x - 7)/2 ) = x - 8

NB. (x - 7)/2 replaces x in f^-1(x).

... (gf)^-1(x) = x - 8

Alternatively

As f(x) = (x + 1)/2 and g : x ---> 2x + 7

... gf(x) = g( (x + 1)/2 )

= 2 x ( (x + 1)/2 ) + 7 = x + 8

Finding the inverse

... y = x + 8
... x = y - 8 Interchanging x for y
... y = x - 8
... (gf)^-1 (x) = x - 8

We will now begin to review coordinate 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. From the diagram, the coordinates of the mid-point are:

M = (x₂ + x_1)/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 co-ordinates.

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

... m = -2/3

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

Graduates and tutors of Revan's University and International Management Centres Association, MSc. Hospitality and Tourism Class of 2008 after their graduation at Half Moon in Montego Bay on last year. Sharing in the occasion was Carolle Guntley (right), director general in the Ministry of Tourism. - photo by Janet Silvera

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