Word Count : 493

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Let's analyze the function \( f(x) = \frac{x^2}{1 + x^2} \). ### Injectivity: A function is injective (one-to-one) if different inputs map to different outputs, i.e., if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). For this function, let's assume: \[ f(x_1) = f(x_2) \] \[ \frac{x_1^2}{1 + x_1^2} = \frac{x_2^2}{1 + x_2^2} \] Multiply both sides by \( (1 + x_1^2)(1 + x_2^2) \): \[ x_1^2(1 + x_2^2) = x_2^2(1 + x_1^2) \] Simplifying: \[ x_1^2 + x_1^2 x_2^2 = x_2^2 + x_1^2 x_2^2 \] The \( x_1^2 x_2^2 \) terms cancel out, leaving: \[ x_1^2 = x_2^2 \] This implies that: \[ x_1 = \pm x_2 \] Thus, the function is not injective, because \( f(x) = f(-x) ______ ________ ______ __________ ____ _______ __________ ______.
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