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Example 29 - Find all points of local maxima and local minima

Example 29 - Chapter 6 Class 12 Application of Derivatives - Part 2

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Example 29 - Chapter 6 Class 12 Application of Derivatives - Part 3

Example 29 - Chapter 6 Class 12 Application of Derivatives - Part 4
Example 29 - Chapter 6 Class 12 Application of Derivatives - Part 5

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Transcript

Example 29 (Method 1) Find all points of local maxima and local minima of the function f given by 𝑓 (π‘₯)=π‘₯3 – 3π‘₯ + 3. 𝑓(π‘₯)=π‘₯3 – 3π‘₯+3 Finding 𝒇^β€² (𝒙) 𝑓′(π‘₯)= 3π‘₯^2 – 3+0 𝑓′(π‘₯)= 3(π‘₯^2βˆ’1) Putting 𝒇′(𝒙)= 0 3(π‘₯^2βˆ’1)=0 π‘₯^2βˆ’1=0 (π‘₯βˆ’1)(π‘₯+1)=0 Thus 𝒙 = 1 , –1 are only critical points Finding maximum & minimum value of 𝑓(π‘₯)=π‘₯3 – 3π‘₯+3 Minimum value 𝑓(1)=(1)3 –3(1)+3= 1 – 3 + 3 = 1 Maximum value 𝑓(βˆ’1)=(βˆ’1)3 –3(βˆ’1)+3= –1 +3 + 3 = 5 Example 29 (Method 2) Find all points of local maxima and local minima of the function f given by 𝑓 (π‘₯) = π‘₯3 – 3π‘₯ + 3. 𝑓(π‘₯)=π‘₯3 – 3π‘₯+3 Finding 𝒇^β€² (𝒙) 𝑓′(π‘₯)= 3π‘₯^2 – 3+0 𝑓′(π‘₯)= 3(π‘₯^2βˆ’1) Putting 𝒇′(𝒙)= 0 3(π‘₯^2βˆ’1)=0 π‘₯^2βˆ’1=0 (π‘₯βˆ’1)(π‘₯+1)=0 So, x = 1 & x = βˆ’1 Finding 𝒇′′(𝒙) 𝑓^β€² (π‘₯)=3(π‘₯^2βˆ’1) 𝑓^β€²β€² (π‘₯)=3 𝑑(π‘₯^2 βˆ’ 1)/𝑑π‘₯ 𝑓^β€²β€² (π‘₯)=3(2π‘₯βˆ’0) 𝑓^β€²β€² (π‘₯)=6π‘₯ Putting 𝒙=βˆ’πŸ 𝑓^β€²β€² (π‘₯) = 6(βˆ’1) <0 π‘₯ = –1 is point of local maxima Putting 𝒙=𝟏 𝑓^β€²β€² (π‘₯) = 6(1) >0 π‘₯ = 1 is point of local minima Finding maximum & minimum value of 𝑓(π‘₯)=π‘₯3 – 3π‘₯+3 Minimum value at x = 1 𝑓(1)=(1)3 –3(1)+3= 1 – 3 + 3 = 1 Maximum value at x = βˆ’1 𝑓(βˆ’1)=(βˆ’1)3 –3(βˆ’1)+3= –1 + 3 + 3 = 5

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.