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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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 โˆ†๐‘ฅ=0.02 Finding fโ€™(x) fโ€™(x) = 6x + 5 Now, โ–ณy = fโ€™(x) โˆ†๐’™ = (6x + 5) 0.02 Putting x = 3 = (6 ร— 3 + 5) 0.02 = 23 ร— 0.02 = 0.46 Now, f(x + โˆ†๐’™) = f(x) + โˆ†๐’š f(3.02) = f(3) + 0.46 = (3 ร— 32 + 5 ร— 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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