Check sibling questions

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

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

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


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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.