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Ex 6.4, 2 - Find approximate value f(2.01), f(x) = 4x2+5x+2 - Finding approximate value of function

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Ex 6.4,2 Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2. Let x = 2 and ∆ x = 0.01 Given f(x) = 4x2 + 5x + 2 f’(x) = 8x + 5 Now, ∆𝑦 = f’(x) ∆ x = (8x + 5) 0.01 Also, ∆y = f (x + ∆x) − f(x) f(x + ∆ x) = f (x) + ∆𝑦 f (2.01) = 4x2 + 5x + 2 + (8x + 5)(0.01) Putting value of x, ∆𝑥 & ∆𝑦 𝑓 2.01﷯=4 2﷯﷮2﷯+ 5 2﷯+2+ 0.01﷯ 8 2﷯+5﷯ = 16+10+2﷯+ 21﷯ 0.01﷯ = 28+0.21=28.21 Hence, the approximate value of f (2.01) is 28.21

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