Functions-11

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Let f : R → R be a function given by f(x) = x2 + 1. Find
(i)f-1 {-5}
(ii) f-1 {26}
(iii) f-1 {10, 37}.

Solution: Recall that if f : A → B such that y ∈ B. Then f-1 {y} = {x ∈ A : f(x) = y}. In other words,

f-1 {y} is the set of pre-images of y.

(i) Let f-1 {-5} = x. Then f(x) = -5 => x2 + 1 = -5

Clearly, this equation is not solvable in R. So, f-1 {-5} = Φ.

(ii) Let f-1 {26} = x. Then f(x) = 26 => x2 + 1 = 26 => x = ± 5.

Therefore, f-1 {26} = {-5, 5}.

(iii) Let f-1 {10, 37} = x. Then f(x) = 10 or f(x) = 37.

Therefore, x2 + 1 = 10 or, x2 + 1 = 37 => x = ± 3 or, x = ±6

So, f-1 {10, 37} = {3, -3, 6, -6}

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