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The function f:R->[-1/2,1/2] defined as ...

The function `f:R->[-1/2,1/2]` defined as `f(x)=x/(1+x^2)` is

A

neither injective nor surjective.

B

invertible.

C

injective but not surjective.

D

Surjective but not injective.

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To determine the properties of the function \( f: \mathbb{R} \to \left[-\frac{1}{2}, \frac{1}{2}\right] \) defined as \( f(x) = \frac{x}{1+x^2} \), we will analyze whether the function is injective (one-to-one), surjective (onto), and whether it is invertible. ### Step 1: Check if the function is injective A function is injective if different inputs produce different outputs. To check this, we assume \( f(x_1) = f(x_2) \) for \( x_1, x_2 \in \mathbb{R} \) and see if it implies \( x_1 = x_2 \). 1. Start with the equation: \[ f(x_1) = f(x_2) \implies \frac{x_1}{1+x_1^2} = \frac{x_2}{1+x_2^2} \] 2. Cross-multiply: \[ x_1(1 + x_2^2) = x_2(1 + x_1^2) \] 3. Expand both sides: \[ x_1 + x_1 x_2^2 = x_2 + x_2 x_1^2 \] 4. Rearranging gives: \[ x_1 - x_2 + x_1 x_2^2 - x_2 x_1^2 = 0 \] 5. Factor out \( x_1 - x_2 \): \[ x_1 - x_2 + (x_1 x_2^2 - x_2 x_1^2) = 0 \implies (x_1 - x_2)(1 - x_1 x_2) = 0 \] 6. This implies either \( x_1 = x_2 \) or \( x_1 x_2 = 1 \). Since \( x_1 x_2 = 1 \) can produce distinct values of \( x_1 \) and \( x_2 \) (e.g., \( x_1 = 2 \) and \( x_2 = \frac{1}{2} \)), the function is not injective. ### Step 2: Check if the function is surjective A function is surjective if every element in the codomain has a pre-image in the domain. 1. We need to find the range of \( f(x) = \frac{x}{1+x^2} \). 2. Set \( y = f(x) \): \[ y = \frac{x}{1+x^2} \] 3. Rearranging gives: \[ y(1 + x^2) = x \implies y + yx^2 - x = 0 \] 4. This is a quadratic equation in \( x \): \[ yx^2 - x + y = 0 \] 5. The discriminant \( D \) must be non-negative for \( x \) to have real solutions: \[ D = (-1)^2 - 4(y)(y) = 1 - 4y^2 \geq 0 \] 6. Solving \( 1 - 4y^2 \geq 0 \) gives: \[ 4y^2 \leq 1 \implies |y| \leq \frac{1}{2} \implies -\frac{1}{2} \leq y \leq \frac{1}{2} \] 7. Since the range of \( f(x) \) is exactly \( \left[-\frac{1}{2}, \frac{1}{2}\right] \), which matches the codomain, the function is surjective. ### Step 3: Conclusion - The function \( f(x) = \frac{x}{1+x^2} \) is **not injective** (as shown in Step 1) and **is surjective** (as shown in Step 2). - Therefore, the function is **surjective but not injective**. ### Final Answer The function \( f(x) = \frac{x}{1+x^2} \) is **surjective but not injective**. ---

To determine the properties of the function \( f: \mathbb{R} \to \left[-\frac{1}{2}, \frac{1}{2}\right] \) defined as \( f(x) = \frac{x}{1+x^2} \), we will analyze whether the function is injective (one-to-one), surjective (onto), and whether it is invertible. ### Step 1: Check if the function is injective A function is injective if different inputs produce different outputs. To check this, we assume \( f(x_1) = f(x_2) \) for \( x_1, x_2 \in \mathbb{R} \) and see if it implies \( x_1 = x_2 \). 1. Start with the equation: \[ ...
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