InterviewSolution
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InterviewSolution
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Find ∆f and df for the function f for the indicated values of x, ∆x and compare (i) f (x) = x3 – 2x2; x = 2, ∆x = dx = 0.5 (ii) f(x) = x2 + 2x + 3; x = -0.5, ∆x = dx = 0.1 |
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Answer» (i) y = f(x) = x3 – 2x2 dy = (3x2 – 4x) dx dy (when x = 2 and dx = 0.5) = [3(22) – 4(2)] (0.5) = (12 – 8)(0.5) = 4(0.5) = 2 (i.e.,) df = 2 Now ∆f = f(x + ∆x) – f(x) Here x = 2 and ∆x = 0.5 f(x) = x3 – 2x2 So f(x + ∆x) = f(2 + 0.5) = f(2.5) = (2.5)3 – 2(2.5)2 = (2.5)2 [2.5 – 2] = 6.25 (0.5) = 3.125 f(x) = f(2) = 23 – 2(22) = 8 – 8 = 0 So ∆f = 3.125 – 0 = 3.125 (ii) y = f(x) = x2 + 2x + 3 dy = (2x + 2) dx dy (when x = – 0.5 and dx = 0.1) = [2(-0.5) + 2] (0.1) = (-1 + 2) (0.1) = (1) (0.1) = 0.1 (i.e.,) df = 0.1 Now ∆f = f(x + ∆x) – f(x) Here x = -0.5 and ∆x = 0.1 x2 + 2x + 3 f(x + ∆x) = f(-0.5 + 0.1) = f(-0.4)2 = (-0.4) + 2(-0.4) + 3 = 0.16 – 0.8 + 3 = 3.16 – 0.8 = 2.36 f(x) = f(-0.5) = (-0.5)2 + 2(-0.5) + 3 = 0.25 – 1 + 3 = 3.25 – 1 = 2.25 So ∆ f = f(x + ∆x) – f(x) = 2.36 – 2.25 = 0.11 |
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