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

Example 23 - Chapter 6 Class 12 Application of Derivatives - Part 2

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Transcript

Question 10 Find the approximate value of f (3.02), where f (x) = 3x2 + 5x + 3.Given f(x) = 3x2 + 5x + 3 where x = 3 and ∆𝑥=0.02 Finding f’(x) f’(x) = 6x + 5 Now, △y = f’(x) ∆𝒙 = (6x + 5) 0.02 Putting x = 3 = (6 × 3 + 5) 0.02 = 23 × 0.02 = 0.46 Now, f(x + ∆𝒙) = f(x) + ∆𝒚 f(3.02) = f(3) + 0.46 = (3 × 32 + 5 × 3 + 3) + 0.46 = (27 + 15 + 3) + 0.46 = 45 + 0.46 = 45.46 Hence, approximate value of f(3.02) is 45.46

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.