Show that the function f: Q → Q, defined by f(x) = 3x + 5, is invertible. Also, find f−1

Asked by Aaryan | 1 year ago |  91

##### Solution :-

Given function f: Q → Q, defined by f(x) = 3x + 5

Now we have to show that 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 = $$\dfrac{ (y -5)}{3}$$ belongs to Q domain

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Now we have to find f-1:

Let f-1(x) = y…… (1)

⇒ x = f(y)

⇒ x = 3y + 5

⇒ x −5 = 3y

⇒ y =$$\dfrac{ (x -5)}{3}$$

Now substituting this value in (1) we get

So, f-1(x) = $$\dfrac{ (x -5)}{3}$$

Answered by Sakshi | 1 year ago

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