f(x) and g(x) are polynomials.

⇒ f: R → R and g: R → R.

So, fog: R → R and gof: R → R.

**(i)** (fog) (x) = f (g (x))

= f (x^{2} + 1)

= 2 (x^{2 }+ 1) + 5

=2x^{2} + 2 + 5

= 2x^{2} +7

**(ii) **(gof) (x) = g (f (x))

= g (2x +5)

= (2x + 5)^{2} + 1

= 4x^{2} + 20x + 26

**(iii)** (fof) (x) = f (f (x))

= f (2x +5)

= 2 (2x + 5) + 5

= 4x + 10 + 5

= 4x + 15

**(iv)** f^{2} (x) = f (x) x f (x)

= (2x + 5) (2x + 5)

= (2x + 5)^{2}

= 4x^{2} + 20x +25

Hence, from (iii) and (iv) clearly fof ≠ f^{2}

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?