A function f: R → R is defined as f(x) = x3 + 4. Is it a bijection or not? In case it is a bijection, find f−1 (3).

Asked by Aaryan | 1 year ago |  181

##### Solution :-

Given that f: R → R is defined as f(x) = x3 + 4

Injectivity of f:

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

Such that f (x) = f (y)

⇒ x3 + 4 = y3 + 4

⇒ x3 = y3

⇒ x = y

So, f is one-one.

Surjectivity of f:

Let y be in the co-domain (R),

Such that f(x) = y.

⇒ x3 + 4 = y

⇒ x3 = y – 4

⇒ x =$$3\sqrt{ (x-4)}$$ in R (domain)

⇒ f is onto.

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

Finding f-1:

Let f−1 (x) = y…… (1)

⇒ x = f (y)

⇒ x = y+ 4

⇒ x − 4 = y3

⇒ y =$$3\sqrt{ (x-4)}$$

So, f-1(x) =$$3\sqrt{ (x-4)}$$  [from (1)]

f-1 (3) = $$3\sqrt{ (x-4)}$$

=$$3\sqrt{-1}$$

= -1

Answered by Sakshi | 1 year ago

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