Back to: Further Mathematics SS1
COMPOSITE MAPPING:
A mapping is composite when the co- domain of the first mapping is the domain of the second mapping.
Consider the mapping f;X→ Z and g: Z→Y
SOLUTION:
(a) F(-3) + F(4) = -5 + 9 = 4
(b) F(2) + F(-5) = 5 +(-10) = -5
(c) G[f(-3)] = g(-5) = -10
(d) G[f(-3)] + g[f(4)] = g(-5) + g(9) = -10 + 18 = 8
Example 2: The functions f and g on the set of real numbers are defined by f(x) = 3x-1 and g(x) = 5x+2 respectively. Find (a) F [g(x)] (b) g [f(x)] (c) 2f(x) – g(x)
SOLUTION:
(a) f[g(x)] = f(5x+2), 5x+2 will represent x in f(x)
f ( 5x +2) = 3 (5x+2) – 1
= 15x +5
(b) g [f(x)] = g( 3x-1)
= g(3x-1) = 5(3x-1) + 2
= 15x -5 +2
= 15x-3
(c) 2f(x) – g(x) = 2(3x-1) – (5x+2)
= 6x -2 -5x -2
= x-4
INVERSE MAPPING:
A function has an inverse if it’s both one- one and onto. Consider the function f(x) = x-3 on the set p={ -1, 5, 9} into set
Q ={ -2,1,3 }
Example: The function f is defined on the ser of real numbers by f(x) = 2x-1, (x≠-2/3)
3x +2
Determine (a) f-1(x) (b) f-1 (-2) (c) determine the largest domain of f-1(x)
SOLUTION:
(a) F(x) = 2x-1
3x+2
f-1(x), y = 2x-1
3x+2
(3x+2) y = 2x-1
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