Classify the following functions f(x) defined in `R to R` as injective , surjective , both or none .
`f(x) = x^(2)`

Answer

To classify the function \( f(x) = x^2 \) defined from \( \mathbb{R} \) to \( \mathbb{R} \) as injective, surjective, both, or neither, we will analyze its properties step by step.

### Step 1: Determine if the function is injective (one-to-one)

A function is injective if different inputs produce different outputs. In mathematical terms, \( f(a) = f(b) \) implies \( a = b \).

For our function:
\[ f(x) = x^2 \]
Let’s check if \( f(a) = f(b) \):
\[ a^2 = b^2 \]
This implies:
\[ a = b \quad \text{or} \quad a = -b \]
Since \( a \) can be equal to \( b \) or \( -b \), we can see that two different values of \( x \) (for example, \( 1 \) and \( -1 \)) yield the same output:
\[ f(1) = 1^2 = 1 \quad \text{and} \quad f(-1) = (-1)^2 = 1 \]
Thus, \( f(x) = x^2 \) is **not injective**.

### Step 2: Determine if the function is surjective (onto)

A function is surjective if every element in the codomain (in this case, \( \mathbb{R} \)) is the image of at least one element from the domain.

The range of \( f(x) = x^2 \) is:
\[ \text{Range} = [0, \infty) \]
This means that the function can only produce non-negative outputs. Therefore, there are no \( x \) values such that \( f(x) \) gives a negative output.

Since the codomain is \( \mathbb{R} \) (which includes negative numbers), and there are no \( x \) values that can produce negative outputs, the function is **not surjective**.

### Conclusion

Since \( f(x) = x^2 \) is neither injective nor surjective, we classify it as **neither**.

### Summary of Classification

- **Injective**: No
- **Surjective**: No
- **Classification**: Neither

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