
Ex 6.4
Ex 6.4, 1 (ii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (iii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (iv) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (v) Important Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (vi) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (vii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (viii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (ix) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (x) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (xi) Important Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (xii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (xiii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (xiv) Important Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (xv) Deleted for CBSE Board 2022 Exams
Ex 6.4,2 Deleted for CBSE Board 2022 Exams
Ex 6.4,3 Important Deleted for CBSE Board 2022 Exams You are here
Ex 6.4,4 Deleted for CBSE Board 2022 Exams
Ex 6.4,5 Important Deleted for CBSE Board 2022 Exams
Ex 6.4,6 Deleted for CBSE Board 2022 Exams
Ex 6.4,7 Deleted for CBSE Board 2022 Exams
Ex 6.4,8 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 6.4,9 (MCQ) Deleted for CBSE Board 2022 Exams
Last updated at April 15, 2021 by Teachoo
Ex 6.4, 3 Find the approximate value of f (5.001), where f (x) = x3 â 7x2 + 15.Let x = 5 and â x = 0.001 Given f (x) = x3 â 7x2 + 15 đâ(x) = 3x2 â 14x Now, ây = fâ(x) âđĽ = (3x2 â 14x) 0.001 Also, ây = f (x + âx) â f(x) f(x + âx) = f(x) + ây f (5.001) = x3 â 7x2 + 15 + (3x2 â 14x) 0.001 Putting value of x = 5 f (5.001) = 53 â 7(5)2 + 15 + (0.001) [3"(5)2" â14(5)] = (125 â 175 + 15) + (0.001) (5) = â35 + 0.005 = â34.995 Hence, the approximate value of f (5.001) is â34.995