text: Q. 2
  
  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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