Prove that the function f: [0, ∞) → R given by

f(x) = 9x² + 6x – 5 is not invertible. Modify the codomain of the function f to make it invertible, and hence find f^-1.

OR

Check whether the relation R in the set R of real numbers, defined by R = {(a, b) : 1 + ab > 0}, is reflexive, symmetric or transitive.

f is invertible if it is one-one as well as onto.

f(x) = 9x² + 6x – 5

⇒ f’(x) = 18x + 6 > 0 for all x > 0

So, f(x) is one-one.

Now, let y R, then for any x,

f(x) = y if y = 9x² + 6x – 5

⇒ y = (3x)² + 2(3x) (1) + (1)² – 5 – 1²

⇒ y = (3x+1)² – 6

As x[0,∞) which means x is a positive real number.

So, x cannot be equal to

Now, fory = - 6

does not belongs to[0,∞)

Hence, f(x) is not onto

⇒ f(x) is not invertible

Now procedure to make it invertible.

Since, x ≥ 0, therefore

Redefining, f: [0, ∞ ) → [ –5, ∞) makes f onto

Now, checking if f: [0, ∞ ) → [ –5, ∞) is one – one as well as onto.

f(x) = 9x² + 6x – 5

⇒ f’(x) = 18x + 6

⇒ f’(x) > 0 for all x [0, ∞ )

So, f(x) is one-one in [0, ∞ ) … (1)

As, f(x) is an increasing function.

As, x [0, ∞ )

So, minimum value of f(x) is at x = 0,

f(0) = 9(0) + 6(0) – 5

⇒ f (0) = -5

Also, as x → ∞, f(x) → ∞.

Also, f(x) is continuous for all x, so f(x) attains all values in

[–5, ∞).

range of f(x) = co-domain of f(x).

So, f is onto. … (2)

From (1) and (2)-

f(x) is bijective.

Hence f is invertible and f^-1: [-5, ∞) → [0, ∞)

OR

Reflexive:

R is reflexive, as 1 + a.a = 1 + a² > 0

Because square of a number is always positive and 1 added to a positive number is a positive number.

⇒ (a, a) R ∀ aR

Symmetric:

If (a, b) R then, 1 + ab> 0

⇒ 1 + ba> 0 ⇒ (b, a) R

Hence, R is symmetric.

Transitive:

Let a=-8, b=-1,

Since, 1 + ab = 1 + (–8) (–1) = 9> 0

∴ (a, b) R

also,

∴ (b, c) R

but,

Hence, R is not transitive.