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Functions
Clement Radcliffe, Contributor
 |
| A
very diligent student from St. Hugh's High School reads a book during an ISSA/KFC
Schoolboy Basketball match at the Stadium Courts, on Tuesday, January 23. - Photo
by Anthony Minott | 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 given last week.
| Prove
that if g: x 2x- 1 then g-1
is | x
+ 1 | | | 2 |
Solution
Since
g:x 2x - 1, then g(x) = y = 2x - 1 ie.
y =
ie.
2x = | 2x
- 1.
y +
1 | | ie.
x = | y
+ 1 | | | 2
|
Interchanging
x for y
If
f and g are defined as follows: f:x
3x-5 and g:x 1/2x (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 (g f)-1
(x). Solution
(a)
Since f : x 3x - 5, then f (x) = 3x - 5 ie.
f (3) = 3 x 3 - 5 = 4 f (3) = 4. (b)
(i) f(x) = 3x - 5, then y = 3x - 5 | ie.3x
= y + 5 x = | y
+ 5 | | | 3 |
Interchanging
x for y
(ii)
I am sure that by using a similar approach you can show that g - 1 (x) = 2x (c)
Given that g(x) = x/2 and f(x) = 3x -5 then
gf(x) = g(3x - 5). N.B. 3x - 5 replaces x in g(x).
ie. 2y = 3x - 5 Interchanging
x for y | ie.
y = | 2x
+ 5 | | ie.
[gf] -1(x) = | 2x
+ 5 | | | 3 | | | 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.
Gradients
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)t hen the formula is: Example:
Find
the gradient m of the line joining the points A(2 , 5) , B(I , 2). 
Mid-point
This
point is denoted by M and from the diagram, the coordinates of the mid-point are:
Example:
Find
the coordinates of M, mid-point of A(2 , 5) and B( 1 , 2). Substituting
into the formula above: Answer:
3/2 , 7/2 In
review 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
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. Clement
Radcliffe is the principal of Glenmuir High School in May Pen. | 
