Ex 6.4, 3 - Find approx value f(5.001), f(x) = x3 - 7x2 + 15 - Finding approximate value of function

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  1. Chapter 6 Class 12 Application of Derivatives
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Ex 6.4,3 Find the approximate value of f (5.001), where f (x) = x3 โ€“ 7x2 + 15. Let x = 5 and โˆ† x = 0.001 Given f (x) = x3 โ€“ 7x2 + 15 ๐‘“โ€™(x) = 3x2 โˆ’ 14x Now, โˆ†y = fโ€™(x) โˆ†๐‘ฅ = (3x2 โˆ’ 14x) 0.001 Also, โˆ†y = f (x + โˆ†x) โˆ’ f(x) f(x + โˆ†x) = f(x) + โˆ†y f (5.001) = x3 โˆ’ 7x2 + 15 + (3x2 โˆ’ 14x) 0.001 Putting value of x, โˆ†๐‘ฅ & โˆ†๐‘ฆ f (5.001) = 53 โˆ’ 7(5)2 + 15 + (0.001) 3(5)2โˆ’14(5)๏ทฏ = (125 โˆ’ 175 + 15) + (0.001) (5) = โˆ’ 35 + 0.005 = โˆ’34.995 hence, the approximate value of f (5.001) is โˆ’34.995

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