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Example 33 - Find all points of local maxima, minima - NCERT

Example 33 - Chapter 6 Class 12 Application of Derivatives - Part 2

Example 33 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 33 - Chapter 6 Class 12 Application of Derivatives - Part 4 Example 33 - Chapter 6 Class 12 Application of Derivatives - Part 5


Transcript

Example 33 (Method 1) Find all the points of local maxima and local minima of the function f given by 𝑓(π‘₯)=2π‘₯3 –6π‘₯2+6π‘₯+5. 𝑓(π‘₯)=2π‘₯3 –6π‘₯2+6π‘₯+5 Finding f’(𝒙) 𝑓’(π‘₯)=𝑑(2π‘₯3 βˆ’ 6π‘₯2 + 6π‘₯ + 5)/𝑑π‘₯ 𝑓’(π‘₯)=6π‘₯^2βˆ’12π‘₯+6+0 𝑓’(π‘₯)=6(π‘₯^2βˆ’2π‘₯+1) 𝑓’(π‘₯)=6((π‘₯)^2+(1)^2βˆ’2(π‘₯)(1)) 𝑓’(π‘₯)=πŸ”(π’™βˆ’πŸ)^𝟐 Putting f’(𝒙)=𝟎 6(π‘₯βˆ’1)^2=0 (π‘₯βˆ’1)^2=0 π‘₯βˆ’1=0 So, 𝒙=𝟏 is only critical point Hence 𝒙=𝟏 is point of inflexion Example 33 (Method 2) Find all the points of local maxima and local minima of the function f given by 𝑓(π‘₯)=2π‘₯3 –6π‘₯2+6π‘₯+5. 𝑓(π‘₯)=2π‘₯3 –6π‘₯2+6π‘₯+5 Finding f’(𝒙) 𝑓’(π‘₯)=𝑑(2π‘₯3 βˆ’ 6π‘₯2 + 6π‘₯ + 5" " )/𝑑π‘₯ 𝑓’(π‘₯)=6π‘₯^2βˆ’12π‘₯+6++0 𝑓’(π‘₯)=6(π‘₯^2βˆ’2π‘₯+1) 𝑓’(π‘₯)=6((π‘₯)^2+(1)^2βˆ’2(π‘₯)(1)) 𝑓’(π‘₯)=πŸ”(π’™βˆ’πŸ)^𝟐 Putting f’(𝒙)=𝟎 6(π‘₯βˆ’1)^2=0 (π‘₯βˆ’1)^2=0 So, 𝒙=𝟏 is the only critical point Finding f’’(𝒙) f’’(π‘₯)=6.(𝑑(π‘₯ βˆ’ 1)^2)/𝑑π‘₯ f’’(π‘₯)=6 Γ— 2(π‘₯βˆ’1) f’’(π‘₯) = 12 (π‘₯βˆ’1) Putting 𝒙=𝟏 f’’(π‘₯)=12(1βˆ’1) = 0 Since f’’(1) = 0 Hence, π‘₯=1 is neither point of Maxima nor point of Minima ∴ 𝒙=𝟏 is Point of Inflexion.

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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.