Example 23 - Find approximate value of f(3.02), f(x)=3x2+5x+3 - Examples

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  1. Chapter 6 Class 12 Application of Derivatives
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Example 23 Find the approximate value of f (3.02), where f (x) = 3x2 + 5x + 3. Let x = 3 and ∆𝑥=0.02 Given f(x) = 3x2 + 5x + 3 f’(x) = 6x + 5 Now, △y = f’(x) ∆𝑥 = (6x + 5) 0.02 Also, ∆𝑦 = f(x + ∆𝑥) − f(x) f (x + ∆𝑥) = f(x) + ∆𝑦 f (3.02) = (3x2 + 5x + 3) + (6x + 5) 0.02 Putting value of x, ∆𝑥 & ∆𝑦 f (3.02) = ﷐3﷐﷐3﷯﷮2﷯5﷐3﷯+3﷯+﷐6﷐3﷯+5﷯0.02 = (27 + 15 + 3) + (23) 0.02 = 45 + 0.46 = 45.46 Hence, approximate value of f (3.02) is 45.46

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