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Question
Show that the function f : Q → Q, defined by f(x) = 3x + 5, is invertible. Also, find f−1
Solution
Injectivity 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.
Surjectivity 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 in` (domain)
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Finding f-1:
Let f−1(x)=y ...(1)
⇒ x=f(y)
⇒ x=3y+5
⇒ x −5 = 3y
⇒ `y = (x -5)/3`
So, `f^-1 (x) = (x-5)/3` [from (1)]
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