MILD-926 JAV Idola Jepang: Sakura Kusumoto Menginspirasi)data {?> <|endoftext|>Human: Given the definition of the function f: R → R where f(x) = x^2 + 1, demonstrate that the function f is not injective (one-to-one) by finding two distinct real numbers x1 and x2 such that f(x1) = f(x2). Assistant: To demonstrate that the function ( f: mathbb{R} o mathbb{R} ) defined by ( f(x) = x^2 + 1 ) is not injective, we need to find two distinct real numbers ( x_1 ) and ( x_2 ) such that ( f(x_1) = f(x_2) ). Let's choose ( x_1 = 1 ) and ( x_2 = -1 ). Now, we need to check if ( f(x_1) = f(x_2) ). First, calculate ( f(1) ): [ f(1) = 1^2 + 1 = 1 + 1 = 2 ] Next, calculate ( f(-1) ): [ f(-1) = (-1)^2 + 1 = 1 + 1 = 2 ] Since ( f(1) = f(-1) = 2 ), we have found two distinct real numbers ( x_1 = 1 ) and ( x_2 = -1 ) such that ( f(x_1) = f(x_2) ). Therefore, the function ( f(x) = x^2 + 1 ) is not injective. [ oxed{x_1 = 1 ext{ and } x_2 = -1} ] To make the final answer clear: [ oxed{1 ext{ and } -1} ] The function ( f(x) = x^2 + 1 ) is not injective because ( f(1) = f(-1) = 2 ), and ( 1 eq -1 ). - Cuplikan Gratis dan Subtitle Bahasa Indonesia srt.

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