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) = \(
\dfrac{ (x -5)}{3}\)

Let A be the set of first five natural and let R be a relation on A defined as follows: (x, y) R x ≤ y

Express R and R^{-1} as sets of ordered pairs. Determine also

**(i) **the domain of R^{‑1}

**(ii)** The Range of R.

A function f: R → R is defined as f(x) = x^{3} + 4. Is it a bijection or not? In case it is a bijection, find f^{−1} (3).

If f: R → R be defined by f(x) = x^{3} −3, then prove that f^{−1} exists and find a formula for f^{−1}. Hence, find f^{−1 }(24) and f^{−1} (5).

Consider f: R^{+} → [−5, ∞) given by f(x) = 9x^{2} + 6x − 5. Show that f is invertible with f^{-1}(x) = \(
\dfrac{\sqrt{(x+6)-1}}{3}\)

If f(x) = \(\dfrac{ (4x + 3)}{(6x – 4)}\), x ≠ (\( \dfrac{2}{3}\)) show that fof(x) = x, for all x ≠ (\( \dfrac{2}{3}\)). What is the inverse of f?