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Find dy and ∆y of y = f(x) = x2 + x at x = 10 when ∆x = 0.1 |
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Answer» Given f(x) = x2 + x, x = 10 and ∆x = 0.1 dy = f'(x) ∆x = (2x + 1) ∆x = [2(10) + 1] (0.1) = (21) (0.1) = 2.1 ∆y = f(x + ∆x) – f(x) = (x + ∆x)2 + (x + ∆x) – (x2 + x) = x2 + 2x ∆x + (∆x)2 + x + ∆x – x2 – x = 2x ∆x + (∆x)2 + ∆x = 2(10) (0.1) + (0.1)2 + 0.1 = 2 + 0.01 + 0.1 = 2.11 |
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