b. Use the second-order Taylor series to approximate the function f(x) = 0.1 x* – 0.2 x3 – 0.5 x2 – 0.3x + 1.2 at x = 1. Assume that the we only know the value of the function and its derivatives at x = 0 (i.e., f(0) = 1.2). Hint: f(x;+1) = f(x;) + f'(x;)h+"G) h² + Lx) h³ + ...+ 2! 3! n! f'(x) = 0.4x3 – 0.6x2 – x – 0.3, thus f'(0) = -0.3 f"(x) = 1.2x2 - 1.2x - 1, thus f"(0) = -1 = 2.4x – 1.2, thus f"(0) = -1.2

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