Question

8. Letf: R R be defined by f(x) = 3x - 7.
Show that f is invertible.
Find f-1 : R → R.​

Answer 1

Given :

f : R → R , f(x) = 3x - 7

To prove :

f(x) is invertible (ie. one-one and onto)

Proof :

• Whether f(x) is one-one :-

Let f(x1) = f(x2)

=> 3x1 - 7 = 3x2 - 7

=> 3x1 = 3x2

=> x1 = x2

Since f(x1) = f(x2) => x1 = x2 , hence f(x) is one-one function .

• Whether f(x) is onto :-

Let y = f(x)

=> y = 3x - 7

=> 3x = y + 7

=> x = (y + 7)/3 -------(1)

Since the domain of the function is R , thus for x to be real y can be any real number .

=> Range (f) = R

=> Range (f) = Co-domain (f)

Since the range and the Co-domain of the given function are equal , hence f(x) is onto function .

Since the given function f(x) is one-one and as well as onto , thus f(x) is invertible .

Hence proved .

Also ,

If f(x) = y , then

$x = {f}^{ - 1} (y)$ -------(2)

From eq-(1) and (2) , we have ;

${f}^{ - 1} (y) = \frac{y + 7}{3}$

Now ,

Substituting x in place of y , we have ;

${f}^{ - 1} (x) = \frac{x + 7}{3}$

Hence ,

${f}^{ - 1} (x) = \frac{x + 7}{3}$