Q2 of 168 Page 216

Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2.

Let x = 2 and Δx = 0.01. Then, we get,

f(2.01) = f(x + Δx) = 4(x +Δx)2 + 5 ( x + Δx) + 2


Now, Δy = f (x + Δx) – f(x)


⇒ f (x + Δx) = f(x) + Δy


≈ f(x) + f’(x).Δx (as dx = Δx)


⇒ f(2.01) ≈ (4x2 + 5x + 2) + (8x + 5) Δx


= [4(2)2 + 5(2) + 2] +[8(2) + 5] (0.01)


= (16 + 10 + 2) + (16 + 5)(0.01)


= 28 + 0.21


= 28.21


Therefore, the approximate value of f (2.01) is 28.21.


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