Let f : R → R : f(x) = x^{2} + 1. Find

(i) f^{–1} {10}

(ii) f^{–1} {–3}.

Given: f(x) = x^{2} + 1

To find: (i) f^{-1}{10}

We know that, if f: X → Y such that y ∈ Y. Then f^{-1}(y) = {x ∈ X: f(x) = y}.

In other words, f^{-1}(y) is the set of pre – images of y

Let f^{-1}{10} = x. Then, f(x) = 10 …(i)

and it is given that f(x) = x^{2} + 1 …(ii)

So, from (i) and (ii), we get

x^{2} + 1 = 10

⇒ x^{2} = 10 – 1

⇒ x^{2} = 9

⇒ x = √9

⇒ x = ± 3

∴ f^{-1}{10} = {-3, 3}

To find: (ii) f^{-1}{-3}

Let f^{-1}{-3} = x. Then, f(x) = -3 …(iii)

and it is given that f(x) = x^{2} + 1 …(iv)

So, from (iii) and (iv), we get

x^{2} + 1 = -3

⇒ x^{2} = -3 – 1

⇒ x^{2} = -4

Clearly, this equation is not solvable in R

∴ f^{-1}{-3} = ɸ

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