# Show that the function f in A=R-{2/3} defined as - Mathematics

Show that the function f in A=R-{2/3}  defined as f(x)=(4x+3)/(6x-4) is one-one and onto hence find f-1

#### Solution

The given function is

f(x)=(4x+3)/(6x-4)

Let f(x_1)=f(x_2)

(4x_1+3)/(6x_1-4)=(4x_2+3)/(6x_2-4)

⇒ 24x1x2 − 16x­1 + 18x2 − 12 = 24x1x2 + 18x1 − 16x2 − 12

⇒18x2 + 16x2 = 18x1 + 16x1

⇒34x2 = 34x1x1= x2

Therefore f(x) is one − one.

Since, (4x+3)/(6x-4)  is a real number, therefore, for every y in the co–domain (f), there exists a number x in R-{2/3} such that

f(x)=y=(4x+3)/(6x-4)

Therefore, f(x) is onto

Now let y=(4x+3)/(6x-4)

6xy-4y=4x+3

x=(4y+3)/(6y-4)

f^(-1)(x)=(4x+3)/(6x-4)

Is there an error in this question or solution?