11-FUNCTIONS-OPERATIONS-COMPOSITE-INVERSE1.ppt

ALGEBRA OF FUNCTIONS
AND IT’S OPERATIONS
A. SUM
(f + h)(x) = f(x) + h(x)
B. DIFFERENCE
(f – h)(x) = f(x) – h(x)

Example 1:
Let f(x) = x + 1 ; h(x) = 2x2
a.(f+h)(x) = f(x) + h(x)
= (x+1) + (2x2
)
= 2x2
+ x + 1
b. (f-h)(x) = f(x) - h(x)
= (x+1) - (2x2
)
= -2x2
+ x + 1
c. (f∙h)(x) = f(x) · h(x)
= (x+1) · (2x2
)
= 2x3
+ 2x2

Example 2:
Let f(x) = 2x + 1 ; h(x) = 3x
x – 2 x – 2
a.(f+h)(x) = f(x) + h(x)
= 2x+1 + 3x
x-2 x-2
= 5x + 1
x - 2
b. (f-h)(x) = f(x) - h(x)
= 2x+1 - (3x)
x-2 x-2
= 1 - x
x – 2

c. (f∙h)(x) = f(x) · h(x)
= (2x+1) · (3x)
x – 2 x - 2
= 6x2
+ 3x
x2
- 4x + 4

COMPOSITION OF FUNCTION
Given two functions f and h, the
composite function denoted by f ◦ h is
defined by (f ◦ h)(x) = f(h(x))

Example:
Let f(x) = 2x +1 ; h(x) = 3x
1.(f◦h)(1)
h(x) = 3x
h(1) = 3(1)
= 3
f(x) = 2x + 1
f(3) = 2(3) + 1
= 6 + 1
= 7
Therefore, (f◦h)(1) = f(h(1)) = f(3) = 7

Let f(x) = 2x +1 ; h(x) = 3x
2. (h◦f)(1)
f(1) = 2x + 1
= 2(1) + 1
= 2 + 1
= 3
h(x) = 3x
h(3) = 3(3)
= 9
Therefore, (h◦f)(1) = h(f(1)) = h(3) = 9

INVERSE FUNCTION
A function f is one-to-one
ordered pairs in the function have the
same ordinate and different abscissa.
A function f has an inverse
function f-1
if and only if f is one-to-
one.

Example:
1. Let f(x) = 2x -3. Find f-1
(x) given that f is
one-to-one.
Let f(x) = y
Thus, y= 2x-3
Interchange x and y.
x = 2y -3
Solve for y
2y = x + 3
y = x + 3
2
f-1
(x) = x+3
2

2. Let f(x) = 5x. Find the inverse function.
Let f(x) = y
Thus, y= 5x
x = 5y
Solve for y
5y = x
y = x
5
f-1
(x) = x
5

3. Let f(x) = x+10. Find the inverse
function.
Let f(x) = y
Thus, y= x+10
x = y+10
Solve for y
y = x -10
f-1
(x) = x -10

Activity
A. Solve the composite function.
Let f(x) = 2x ; g(x) = 4x2
- 5x ; h(x) = x + 1
1.(g ◦ h)(2)=21
2.(h ◦ g)(1)
3.(f ◦ h)(3)
4.(f ◦ g)(1)
5.(g ◦ f)(1)
6.(h ◦ f)(2)
7.(f ◦ f)(2)
8.(h ◦ h)(3)
9.(g ◦ g)(1)
10.(g ◦ g)(-1)

Solutions
1.h(2)= 2 + 1 = 3 g(3)= 4(9) – 5(3)= 36 – 15 = 21
2.g(1)= 4 – 5 = -1 h(-1)= -1 + 1= 0
3.h(3)= 3 + 1 = 4 f(4)= 2(4) = 8
4.g(1)= 4(1) – 5(1) = -1 f(-1) = 2(-1) = -2
5.f(1)= 2(1) = 2 g(2) = 4(4) – 5(2) = 6
6.f(2)= 2(2) = 4 h(4)= 4 + 1 = 5
7.f(2)= 2(2) = 4 f(4)= 2(4) = 8
8.h(3)= 3 + 1= 4 h(4)= 4 + 1 = 5
9.g(1)= 4 – 5 = -1 g(-1)= 4 + 5 = 9
10.g(-1)= 4 + 5 = 9 g(9)= 4(81) – 45 = 324 – 45 = 279

B. Solve the inverse function.
1.f(x) = 4x – 1
2.f(x) = x – 2
3.f(x) = 2x + 5
4.f(x) = x-2
2x+3
5. f(x) = -3x

Solution
1.x = 4y – 1 4y= x + 1 f-1
= x + 1
4
2.x= y – 2 y= x + 2 f-1
(x)= x + 2
3. x= 2y + 5 2y= x – 5 f-1
(x)= x – 5
2
4.x= y-2 y= 2xy+3x-2 f-1
(x)= 2xy+3x-2
2y+3
5.x= -3yy= x f-1
(x)= x
-3 -3

11-FUNCTIONS-OPERATIONS-COMPOSITE-INVERSE1.ppt

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11-FUNCTIONS-OPERATIONS-COMPOSITE-INVERSE1.ppt