Find the approximate value of f(2.01), where $f (x) = 4 x^{2} + 5 x + 2$ .

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Solution

$C o n s i d e r f (x) = 4 x^{2} + 5 x + 2 \Rightarrow f^{'} (x) = 8 x + 5 L e t x = 2 a n d Δ x = 0.01 A l s o, f (x + Δ x) ≃ f (x) + Δ x f^{'} (x) ∴ f (x + Δ x) ≃ (4 x^{2} + 5 x + 2) + (8 x + 5) Δ x \Rightarrow f (2 + 0.01) ≃ (4 \times 2^{2} + 5 \times 2 + 2) + (8 \times 2 + 5) (0.01) (a s x = 2, Δ x = 0.01) 28 + 21 \times 0.01 = 28.21 \Rightarrow f (2.01) ≃ 28.21$

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