Consider a function `f:R rarr R` defined by `f(x)=x^(3)+4x+5`, then

Options

A. f is one - one but not onto
B. f is onto but not one - one
C. f is one - one and onto
D. f is neither one - one nor onto

Answer

To determine if the function \( f(x) = x^3 + 4x + 5 \) is one-to-one and onto, we will follow these steps:

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

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

1. Assume \( f(x_1) = f(x_2) \):
\[ x_1^3 + 4x_1 + 5 = x_2^3 + 4x_2 + 5 \]
Subtracting 5 from both sides gives:
\[ x_1^3 + 4x_1 = x_2^3 + 4x_2 \]
Rearranging this, we have:
\[ x_1^3 - x_2^3 + 4x_1 - 4x_2 = 0 \]

2. Factor the left-hand side:
\[ (x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2) + 4(x_1 - x_2) = 0 \]
This can be factored as:
\[ (x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2 + 4) = 0 \]

3. For the product to be zero, either:
- \( x_1 - x_2 = 0 \) which gives \( x_1 = x_2 \), or
- \( x_1^2 + x_1x_2 + x_2^2 + 4 = 0 \)

The second equation \( x_1^2 + x_1x_2 + x_2^2 + 4 = 0 \) cannot hold for real numbers since \( x_1^2 + x_1x_2 + x_2^2 \) is always non-negative and adding 4 makes it strictly positive.

Thus, \( f(x) \) is one-to-one.

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

A function is onto if its range is equal to its codomain. Here, the codomain is \( \mathbb{R} \).

1. Analyze the behavior of \( f(x) \) as \( x \) approaches \( -\infty \) and \( +\infty \):
- As \( x \to -\infty \):
\[ f(x) \to -\infty \quad (\text{since } x^3 \text{ dominates}) \]
- As \( x \to +\infty \):
\[ f(x) \to +\infty \quad (\text{since } x^3 \text{ dominates}) \]

2. Since \( f(x) \) is a polynomial function, it is continuous. Therefore, by the Intermediate Value Theorem, \( f(x) \) takes all values between \( -\infty \) and \( +\infty \).

Thus, the range of \( f(x) \) is \( (-\infty, +\infty) \), which matches the codomain \( \mathbb{R} \).

### Conclusion

Since \( f(x) \) is both one-to-one and onto, we conclude that the function \( f(x) = x^3 + 4x + 5 \) is a bijection.

### Final Answer

The function \( f(x) \) is one-to-one and onto.