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Functions
and relations
Clement
Radcliffe, Contributor
²
FOR
THE past two weeks we have been reviewing
functions and relations. You have
been reminded that a function is a
relation in which each element in
the domain is mapped onto one and
only one element in the range. Please
be reminded also that the function
f(x) = 2x - 7 means the function f
such that x is mapped onto 2x - 7.
Let
us now review the solution to the
homework given last week.
1.
* Prove that if g: x  2x
- 1 then g-1
is
SOLUTION
If
g:x
2x - 1
...
g(x) = y = 2x - 1
You
are initially required to express
x in terms of y.
| ...
2x = |
y
+ 1 |
Clearing
|
| ...
x = |
x
+1
|
Interchange
x for y
|
| |
2
|
|
| ...
y = |
x
+1
|
|
| |
2
|
|
| ...
g-1(x)= |
x
+1
|
|
| |
2
|
|
2.
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 (gf)-1
SOLUTION
(a)
Since f:x
3x - 5 ... f(x) = 3x - 5
...
f(3) = 3 x 3 - 5 = 4
...
f(3) = 4
(b)
(i) As f(x) = y = 3x - 5
| ...
3x = |
y
+ 5 |
|
| ...
x = |
x
+5
|
Interchange
x for y
|
| |
3
|
|
| ...
y = |
x
+5
|
|
| |
3
|
|
| ...
f-1(x)= |
x
+5
|
|
| |
3
|
|
(ii)
Since g:x
1/2x ... g(x) = y = 1/2x
| ...
x = |
2y |
Interchange
x for y
|
| ...
y = 2x |
|
|
| ...
g-1(x) |
=
2x
|
|
(c)
Since (gf)-1
(x) = f-1 g-1
(x)
This
relationship has been established
and so may be taken as fact.
From
the above, f(x) = 3x - 5 ... y = 3x
- 5
| ...
3x = |
y
+ 5 |
|
| ...
x = |
x
+5
|
Interchange
x for y
|
| |
3
|
|
| ...
y = |
x
+5
|
|
| |
3
|
|
| ...
f-1(x)= |
x
+5
|
Since
g-1 (x)
= 2x |
| |
3
|
|
| ...
f-1g-1(x)
= f-1 (2x) g-1
replaces x in f(x) |
|
f-1g-1(x)
= |
2x +5
|
|
| |
3
|
|
| [gf]-1(x)
= |
2x
+5
|
|
| |
3
|
|
Alternatively
[gf(x)]-1
may be found by first determining
gf.
Since
f(x) = 3x - 5 and g(x) = 1/2x
Then
gf(x) = g(3x - 5) f(x) replaces x
in g(x)
| ...
gf(x) = |
3x
- 5 |
|
| |
2
|
|
| Let
gf (x) = |
y
=
|
3x
- 5
|
|
| |
|
2
|
|
|
Expressing
x in terms of y
|
| ...
2y = |
3x
- 5
|
|
| ...
3x = |
2y
+ 5
|
|
| ...
x = |
2y
+ 5 |
Interchanging
x for y |
|
|
3
|
|
| [gf]-1(x)
= |
2x
+5
|
|
| |
3
|
|
We
will complete this week's lesson by
working the following examples:
*
Given that f(x) = 2x - 3,
(i)
Determine an expression for f-1(x)
(ii)
Hence, or otherwise, calculate the
value of x for which f-1
(x) = 7.
Question
number 4, CXC General Proficiency,
January 1999.
SOLUTION
(i)
As f(x) = y = 2x - 3
| ...
2x = |
y
+ 3 |
|
| ...
x = |
y
+ 3
|
Interchanging
x for y
|
| |
2
|
|
|
| ...
y = |
x
+ 3
|
|
|
|
|
2
|
|
|
| f-1
(x) = |
x
+ 3
|
|
| |
2
|
|
|
|
|
|
|
(ii)
Since f - 1 (x) = 7
|
|
x
+ 3
|
|
=
7
|
|
|
2
|
|
|
|
| ...
x + 3 = 14 |
| ...
x = 11. |
*
If h(x) = 1 + 3x and k (x) = x +2,
Calculate
(i)
hk(x)
(ii) hk(4)
(iii)
(hk)-1(x)
(iv)
the value of x, when hk(x) = 0.
Question
number 5, CXC General Proficiency,
June 1999.
SOLUTION
(i)
Given that h(x) = 1 + 3x and k(x)
= x +2,
...
hk(x) = h(x + 2) k(x) replaces x in
h(x)
...
hk (x) = 1 + 3 x (x +2) = 1 + 3x +
6
...
hk (x) = 3x + 7
(ii)
Since hk (x) = 3x + 7
...
hk (4) = 3 x 4 + 7 = 19
...
hk (4) = 19
(iii)
Let y = hk (x) = 3x + 7
...
y = 3x + 7
...
3x = y - 7
| ...
x = |
y
- 7 |
Interchanging
x for y
|
| |
3
|
|
| ...
y = |
x
- 7
|
[hk]-1(x)
|
x
- 7 |
| |
3
|
|
3
|
|
|
|
| |
|
|
(iv)
Since hk (x) = 3x + 7 = 0
|
| ...
3x = -7 |
|
3x
- 7
|
|
|
|
3
|
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Next
week, we will begin the review of
vectors. Please attempt some preliminary
work until then.
*
Clement Radcliffe is principal
of Glenmuir High School in Clarendon.
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