Find Range and Pre-Images of \(f(x)=x^2\)

Solution

Given:

$$ f(x)=x^2 $$

(i) Range of \(f\)

Since the square of every real number is always non-negative,

$$ x^2 \ge 0 \quad \text{for all } x \in \mathbb{R} $$

Therefore, the range of \(f\) is:

$$ [0,\infty) $$

(ii) Find \(\{x : f(x)=4\}\)

We have:

$$ f(x)=4 $$

So,

$$ x^2=4 $$

Taking square roots,

$$ x=\pm 2 $$

Hence,

$$ \{x : f(x)=4\}=\{-2,2\} $$

(iii) Find \(\{y : f(y)=-1\}\)

We have:

$$ y^2=-1 $$

But the square of a real number cannot be negative.

Therefore, there is no real number \(y\) such that \(f(y)=-1\).

Hence,

$$ \{y : f(y)=-1\}=\phi $$