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f (x) =4 x 2 + 5x + 2 Let x = 2 and ∆x = 0.01 Now ∆y = f (x + ∆x) – f (x) ∴ f (x + ∆x) = f (x) + ∆y ⇒ f (x + ∆x) = f (x) + f ' (x) ∆x [∵ dx = ∆x] ⇒ (x + ∆x) = f (x) + (8 x + 5) ∆x ∴ f (3.2) = f (2) + {8(2) + 5 } (0.01) = [4 (2) 2 + 5(2)+ 2] + (16 + 5) (0.01) = (16 + 10 + 2) +(21) (0.01) = 28 + 0.21 ∴ f (3.2) = 28.21

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Application of Derivatives

Short Answer Type

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388.
Find the approximate value of f (2.01), where f (x) = 4x² + 5x + 2 .

f (x) =4 x²+ 5x + 2
Let x = 2 and ∆x = 0.01
Now ∆y = f (x + ∆x) – f (x)
∴ f (x + ∆x) = f (x) + ∆y
⇒ f (x + ∆x) = f (x) + f ' (x) ∆x [∵ dx = ∆x]
⇒ (x + ∆x) = f (x) + (8 x + 5) ∆x
∴ f (3.2) = f (2) + {8(2) + 5 } (0.01)
= [4 (2)²+ 5(2)+ 2] + (16 + 5) (0.01)
= (16 + 10 + 2) +(21) (0.01)
= 28 + 0.21
∴ f (3.2) = 28.21

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