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
Example 29 (Method 1) Find all points of local maxima and local minima of the function f given by ๐ (๐ฅ)=๐ฅ3 โ 3๐ฅ + 3. ๐(๐ฅ)=๐ฅ3 โ 3๐ฅ+3 Finding ๐^โฒ (๐) ๐โฒ(๐ฅ)= 3๐ฅ^2 โ 3+0 ๐โฒ(๐ฅ)= 3(๐ฅ^2โ1) Putting ๐โฒ(๐)= 0 3(๐ฅ^2โ1)=0 ๐ฅ^2โ1=0 (๐ฅโ1)(๐ฅ+1)=0 Thus ๐ฅ = 1 , โ1 are only critical points Finding maximum & minimum value of ๐(๐ฅ)=๐ฅ3 โ 3๐ฅ+3 Minimum value ๐(1)=(1)3 โ3(1)+3= 1 โ 3 + 3 = 1 Maximum value ๐(โ1)=(โ1)3 โ3(โ1)+3= โ1 +3 + 3 = 5 Example 29 (Method 2) Find all points of local maxima and local minima of the function f given by ๐ (๐ฅ) = ๐ฅ3 โ 3๐ฅ + 3. ๐(๐ฅ)=๐ฅ3 โ 3๐ฅ+3 Finding ๐^โฒ (๐) ๐โฒ(๐ฅ)= 3๐ฅ^2 โ 3+0 ๐โฒ(๐ฅ)= 3(๐ฅ^2โ1) Putting ๐โฒ(๐)= 0 3(๐ฅ^2โ1)=0 ๐ฅ^2โ1=0 (๐ฅโ1)(๐ฅ+1)=0 So, x = 1 & x = โ1 Finding ๐โฒโฒ(๐) ๐^โฒ (๐ฅ)=3(๐ฅ^2โ1) ๐^โฒโฒ (๐ฅ)=3 ๐(๐ฅ^2 โ 1)/๐๐ฅ ๐^โฒโฒ (๐ฅ)=3(2๐ฅโ0) ๐^โฒโฒ (๐ฅ)=6๐ฅ Putting ๐=โ๐ ๐^โฒโฒ (๐ฅ) = 6(โ1) = โ6<0 ๐ฅ = โ1 is point of local maxima Putting ๐=๐ ๐^โฒโฒ (๐ฅ) = 6(1) = 6>0 ๐ฅ = 1 is point of local minima Finding maximum & minimum value of ๐(๐ฅ)=๐ฅ3 โ 3๐ฅ+3 Minimum value at x = 1 ๐(1)=(1)3 โ3(1)+3= 1 โ 3 + 3 = 1 Maximum value at x = โ1 ๐(โ1)=(โ1)3 โ3(โ1)+3= โ1 + 3 + 3 = 5
Examples
Example 2 Not in Syllabus - CBSE Exams 2021
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Example 4 Important Not in Syllabus - CBSE Exams 2021
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Example 21 Not in Syllabus - CBSE Exams 2021
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Example 42 Important Not in Syllabus - CBSE Exams 2021
Example 43 Important Not in Syllabus - CBSE Exams 2021
Example 44 Important Not in Syllabus - CBSE Exams 2021
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Example 49 Not in Syllabus - CBSE Exams 2021
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