The function `f:R rarr R` defined by `f(x)=x^3-1,` is:

Options

A. One-one but not onto
B. Onto but not one-one
C. A bijection
D. Neither one-one nor onto

Answer

To determine whether the function \( f(x) = x^3 - 1 \) is one-to-one (injective) or onto (surjective), we can analyze the function step by step.

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

To check if the function is one-to-one, we need to see if \( f(x_1) = f(x_2) \) implies that \( x_1 = x_2 \).

1. Start with the equation:
\[ f(x_1) = f(x_2) \]
which means:
\[ x_1^3 - 1 = x_2^3 - 1 \]

2. Simplify the equation:
\[ x_1^3 = x_2^3 \]

3. Take the cube root of both sides:
\[ x_1 = x_2 \]

Since \( f(x_1) = f(x_2) \) leads to \( x_1 = x_2 \), we conclude that the function \( f(x) \) is one-to-one.

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

To check if the function is onto, we need to determine if for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \).

1. Set \( f(x) = y \):
\[ x^3 - 1 = y \]

2. Rearranging gives:
\[ x^3 = y + 1 \]

3. Now, take the cube root:
\[ x = \sqrt[3]{y + 1} \]

Since \( y \) can be any real number, \( y + 1 \) can also be any real number, and thus \( x \) can take any real value. This means that for every \( y \), there exists an \( x \) such that \( f(x) = y \).

Therefore, the function \( f(x) = x^3 - 1 \) is onto.

### Conclusion

Since \( f(x) \) is both one-to-one and onto, we can conclude that \( f(x) \) is a bijective function.

### Final Answer

The function \( f(x) = x^3 - 1 \) is a bijective function.

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