If `f:RtoR,f(x)=x^(2)`, then f is

Options

A. injective but not surjective
B. surjective but not injective
C. bijective
D. None of the above

Answer

To determine the properties of the function \( f: \mathbb{R} \to \mathbb{R}, f(x) = x^2 \), we need to analyze whether it is injective (one-to-one) and surjective (onto).

### Step 1: Check for Injectivity
A function is injective if every element of the range is mapped from a unique element of the domain. This means that if \( f(a) = f(b) \), then \( a \) must equal \( b \).

1. Assume \( f(a) = f(b) \):
\[ a^2 = b^2 \]
2. This implies:
\[ a = b \quad \text{or} \quad a = -b \]
3. Since \( a \) can equal \( b \) or \( -b \), it shows that different inputs can yield the same output (for example, \( f(2) = f(-2) = 4 \)). Thus, the function is **not injective**.

### Step 2: Check for Surjectivity
A function is surjective if every element in the codomain is mapped from at least one element in the domain. Here, the codomain is \( \mathbb{R} \).

1. The function \( f(x) = x^2 \) produces outputs that are always non-negative (i.e., \( f(x) \geq 0 \) for all \( x \in \mathbb{R} \)).
2. Therefore, the range of \( f \) is \( [0, \infty) \).
3. Since the codomain is \( \mathbb{R} \) (which includes negative numbers), not every real number can be achieved as an output of \( f \). Thus, the function is **not surjective**.

### Conclusion
Since the function \( f(x) = x^2 \) is neither injective nor surjective, it is not bijective.

### Final Answer
The function \( f(x) = x^2 \) is **not injective** and **not surjective**.

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