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Question
Find Δf and df for the function f for the indicated values of x, Δx and compare:
f(x) = x3 – 2x2, x = 2, Δx = dx = 0.5
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Solution
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 – (2.5)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
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