Question -
Answer -
Givenfunction f: Q → Q, defined by f(x) = 3x + 5
Now we have to showthat the given function is invertible.
Injection of f:
Let x and y be two elements of the domain (Q),
Such that f(x) = f(y)
⇒ 3x + 5 =3y + 5
⇒ 3x = 3y
⇒ x = y
so, f is one-one.
Surjection of f:
Let y be in the co-domain (Q),
Such that f(x) =y
⇒ 3x +5 = y
⇒ 3x = y – 5
⇒ x = (y -5)/3 belongsto Q domain
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Now we have tofind f-1:
Let f-1(x)= y…… (1)
⇒ x = f(y)
⇒ x = 3y + 5
⇒ x −5 = 3y
⇒ y = (x – 5)/3
Now substituting thisvalue in (1) we get
So, f-1(x)= (x – 5)/3