functions

Asked by  | 4th Oct, 2009, 12:22: PM

Expert Answer:

Given  f(x)= 3x-5/2x+5

For f -1 to exists  f must be one -one and onto

f is one-one

 Let f(x1)=f(x2)

(3x1-5) /(2x1+5)  = (3x2-5)/(2x2+5)

(3x1-5)(2x2+5)=(3x2-5)(2x1+5)

6x1x2+15x1-10x2-25=6x1x2+15x2-10x1-25

So x1 = x2

so f is one-one.

Let y be a real number

s.t f(x)=y

3x-5/2x+5 =y

x = 5y+5/(3-2y) 

which is defined for all y except 3/2

so f-1  =  5x+5/(3-2x) 

Answered by  | 5th Oct, 2009, 10:30: AM

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