Question

f(x)=x²+1 find pre image of 17 and 2.2

Answer 1

Important notes:

Let, f be a mapping that assigns each element of A to B under a definable rule.

We find a f(x) in B for at least one element x in A. Here f(x) is called an image in B of the pre-image x in A.

Solution:

1. When f(x) = 17, x = ?

Here f(x) = x² + 1

Then x² + 1 = 17

or, x² + 1 - 17 = 0

or, x² - 16 = 0

or, x² - 4² = 0

or, (x + 4) (x - 4) = 0

i.e., x = - 4, 4

∴ pre-images of f(x) = 17 are x = - 4, 4

2. When f(x) = 2.2, x = ?

Here f(x) = x² + 1

Then x² + 1 = 2.2

or, x² = 2.2 - 1

or, x² = 1.2

i.e., x = ± √1.2

∴ the pre-images of f(x) = 2.2 are x = ± √1.2

Answer 2

Answer:

Pre-images of f(x) = 17 are x = - 4, 4

Pre-images of f(x) = 2.2 are x = ± √1.2

Step-by-step explanation:

In this question,

We have been given that

f(x) = $\mathbf{x^{2}\ +\ 1}$

We need to find the pre images of 17 and 2.2

1. When f(x) = 17, x = ?

Here f(x) = x² + 1

Then x² + 1 = 17

or, x² + 1 - 17 = 0

or, x² - 16 = 0

or, x² - (4)² = 0               USING [$\mathbf{x^{2}\ -\ y^{2}\ =\ (x\ +\ y)(x\ -\ y)}$]

or, (x + 4) (x - 4) = 0

i.e., x = - 4, 4

Therefore, Pre-images of f(x) = 17 are x = - 4, 4

2. When f(x) = 2.2, x = ?

Here f(x) = x² + 1

Then x² + 1 = 2.2

or, x² = 2.2 - 1

or, x² = 1.2

i.e., x = ± √1.2

Therfore, Pre-images of f(x) = 2.2 are x = ± √1.2