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Question
Find Δf and df for the function f for the indicated values of x, Δx and compare:
f(x) = x2 + 2x + 3, x = – 0.5, Δx = dx = 0.1
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Solution
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)
= (– 0.4)2 + 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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