Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

Last updated at Jan. 7, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
Transcript
Ex 6.4, 2 Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2. Let x = 2 and โ x = 0.01 Given f(x) = 4x2 + 5x + 2 fโ(x) = 8x + 5 Now, โ๐ฆ = fโ(x) โ x = (8x + 5) 0.01 Also, โy = f (x + โx) โ f(x) f(x + โ x) = f (x) + โ๐ฆ f (2.01) = 4x2 + 5x + 2 + (8x + 5)(0.01) Putting value of x=2 ๐ (2.01)=4 (2)^2+ 5(2)+2+(0.01)[8(2)+5] = (16+10+2)+(21) (0.01) = 28+0.21 =28.21 Hence, the approximate value of f (2.01) is 28.21
Ex 6.4
Ex 6.4, 1 (ii) Not in Syllabus - CBSE Exams 2021
Ex 6.4, 1 (iii) Not in Syllabus - CBSE Exams 2021
Ex 6.4, 1 (iv) Not in Syllabus - CBSE Exams 2021
Ex 6.4, 1 (v) Not in Syllabus - CBSE Exams 2021
Ex 6.4, 1 (vi) Not in Syllabus - CBSE Exams 2021
Ex 6.4, 1 (vii) Not in Syllabus - CBSE Exams 2021
Ex 6.4, 1 (viii) Not in Syllabus - CBSE Exams 2021
Ex 6.4, 1 (ix) Not in Syllabus - CBSE Exams 2021
Ex 6.4, 1 (x) Not in Syllabus - CBSE Exams 2021
Ex 6.4, 1 (xi) Not in Syllabus - CBSE Exams 2021
Ex 6.4, 1 (xii) Not in Syllabus - CBSE Exams 2021
Ex 6.4, 1 (xiii) Not in Syllabus - CBSE Exams 2021
Ex 6.4, 1 (xiv) Not in Syllabus - CBSE Exams 2021
Ex 6.4, 1 (xv) Not in Syllabus - CBSE Exams 2021
Ex 6.4,2 Not in Syllabus - CBSE Exams 2021 You are here
Ex 6.4,3 Important Not in Syllabus - CBSE Exams 2021
Ex 6.4,4 Not in Syllabus - CBSE Exams 2021
Ex 6.4,5 Important Not in Syllabus - CBSE Exams 2021
Ex 6.4,6 Not in Syllabus - CBSE Exams 2021
Ex 6.4,7 Not in Syllabus - CBSE Exams 2021
Ex 6.4,8 Not in Syllabus - CBSE Exams 2021
Ex 6.4,9 Not in Syllabus - CBSE Exams 2021
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