Functions-2

Let f : R → R be given by f(x) = x² + 3. Find (a) {x : f(x) = 28} (b) the pre-images of 39 and 2 under f.

Solution: (a) We have f(x) = 28 =>x² + 3 = 28 => x² = 25 => x = ± 5.

Therefore, {x : f(x) = 28} = {-5, 5}.

(b) Let x be the pre-image of 39. Then,

f(x) = 39 => x² + 3 = 39 => x² = 36 => x = ± 6

So, pre-images of 39 are -6 and 6.

Let x be the pre-image of 2. Then,

f(x) = 2 => x² + 3 = 2 => x² = -1

Since, no real value of x satisfies the equation x² = -1

So, 2 does not have any pre-image under f.