Given f: R → R given by f(x) = 4x + 3

Now we have to show that the given function is invertible.

Consider injection of f:

Let x and y be two elements of domain (R),

Such that f(x) = f(y)

⇒ 4x + 3 = 4y + 3

⇒ 4x = 4y

⇒ x = y

So, f is one-one.

Now surjection of f:

Let y be in the co-domain (R),

Such that f(x) = y.

⇒ 4x + 3 = y

⇒ 4x = y -3

⇒ \( \dfrac{y-3}{4}\) in R (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 = 4y + 3

⇒ x − 3 = 4y

⇒ y = \( \dfrac{x-3}{4}\)

Now substituting this value in (1) we get

So, f^{-1}(x) = \(
\dfrac{x-3}{4}\)

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?