The function `f:R to R` given by `f(x)=x^(2)+x` is

Options

A. one-one nad onto
B. one-one and into
C. many-one and onto
D. many one and into

Answer

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

### Step 1: Check if the function is one-to-one
A function is one-to-one if it is either strictly increasing or strictly decreasing. To check this, we will differentiate the function.

**Differentiation:**
\[ f'(x) = \frac{d}{dx}(x^2 + x) = 2x + 1 \]

**Finding critical points:**
Set the derivative equal to zero to find critical points:
\[ 2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2} \]

### Step 2: Analyze the sign of the derivative
We will analyze the sign of \( f'(x) \) around the critical point \( x = -\frac{1}{2} \).

- For \( x < -\frac{1}{2} \):
\[ f'(-1) = 2(-1) + 1 = -2 + 1 = -1 \quad (\text{negative}) \]

- For \( x > -\frac{1}{2} \):
\[ f'(0) = 2(0) + 1 = 0 + 1 = 1 \quad (\text{positive}) \]

### Conclusion on one-to-one
Since \( f'(x) < 0 \) when \( x < -\frac{1}{2} \) and \( f'(x) > 0 \) when \( x > -\frac{1}{2} \), the function is decreasing on \( (-\infty, -\frac{1}{2}) \) and increasing on \( (-\frac{1}{2}, \infty) \). Therefore, the function is not one-to-one.

### Step 3: Check if the function is onto
To check if the function is onto, we need to determine the range of \( f(x) \).

**Finding the minimum value:**
The minimum value occurs at the critical point \( x = -\frac{1}{2} \):
\[ f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \]

### Step 4: Determine the range
Since \( f(x) \) is a quadratic function that opens upwards (the coefficient of \( x^2 \) is positive), the range of \( f(x) \) is:
\[ \text{Range } f(x) = \left[-\frac{1}{4}, \infty\right) \]

### Conclusion on onto
The codomain of \( f \) is \( \mathbb{R} \), but the range is \( \left[-\frac{1}{4}, \infty\right) \). Therefore, the function is not onto.

### Final Conclusion
The function \( f(x) = x^2 + x \) is neither one-to-one nor onto.

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