COMPOSITE MAPPING AND INVERSE MAPPING

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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