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Using differential, find the approximate value of f(2.01), where f(x) = 4x3 + 5x2 + 2.

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Let x = 2,

Δx = 0.01 

Where f(x) = 4x3 + 5x2 + 2

⇒ x + Δx = 2 + 0.01 = 2.01

By definition, approximate value of f(x) is

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