Functions-15

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Let A = {-2, -1, 0, 1, 2} and f : A → Z given by f(a) = x2 - 2x – 3.
Find (a) the range of f (b) pre-images of 6, -3 and 5.

Solution: (a) We have f(-2) = (-2)2 - 2(-2) – 3 = 5, f(-1) = (-1)2 - 2(-1) – 3 = 0,

f(0) = -3, f(1) = 12 - 2 X 1 – 3 = -4 and f(2) = 22 - 2 X 2 – 3= -3.

So range (f) = {0, 5, -3, -4}

(b) Let x be the pre-image of 6. Then,

f(x) = 6 => x2 - 2x – 3 = 6

=>x2 - 2x – 9 = 0

=>x = 1 ± √10.

Since, x = 1 ±10 ∉ A, so, there is no pre-image of 6.

Let x be the pre-image of -3. Then,

f(x) = -3 => x2 - 2x – 3 = -3

=>x2 - 2x = 0 => x = 0, 2.

Clearly, 0, 2 ∈ A. So, 0 and 2 are pre-images of -3.

Let x be the pre-image of 5. Then,

f(x) = 5 => x2 - 2x – 3 = 5 => x2 - 2x – 8 = 0

=>(x - 4)(x + 2) = 0 => x = 4, -2.

Since, -2 ∈ A. So, -2 is the pre-image of 5.

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