Example 21 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 16, 2024 by

Example 21 - Find all points of local maxima, minima - NCERT - Examples

part 2 - Example 21 - Examples - Serial order wise - Chapter 6 Class 12 Application of Derivatives

  part 3 - Example 21 - Examples - Serial order wise - Chapter 6 Class 12 Application of Derivatives

part 4 - Example 21 - Examples - Serial order wise - Chapter 6 Class 12 Application of Derivatives
part 5 - Example 21 - Examples - Serial order wise - Chapter 6 Class 12 Application of Derivatives

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Example 21 (Method 2) Find all the points of local maxima and local minima of the function f given by 𝑓(𝑥)=2𝑥3 –6𝑥2+6𝑥+5. 𝑓(𝑥)=2𝑥3 –6𝑥2+6𝑥+5 Finding f’(𝒙) 𝑓’(𝑥)=𝑑(2𝑥3 − 6𝑥2 + 6𝑥 + 5" " )/𝑑𝑥 𝑓’(𝑥)=6𝑥^2−12𝑥+6++0 𝑓’(𝑥)=6(𝑥^2−2𝑥+1) 𝑓’(𝑥)=6((𝑥)^2+(1)^2−2(𝑥)(1)) 𝑓’(𝑥)=𝟔(𝒙−𝟏)^𝟐 Putting f’(𝒙)=𝟎 6(𝑥−1)^2=0 (𝑥−1)^2=0 So, 𝒙=𝟏 is the only critical point Finding f’’(𝒙) f’’(𝑥)=6.(𝑑(𝑥 − 1)^2)/𝑑𝑥 f’’(𝑥)=6 × 2(𝑥−1) f’’(𝑥) = 12 (𝑥−1) Putting 𝒙=𝟏 f’’(𝑥)=12(1−1) = 0 Since f’’(1) = 0 Hence, 𝑥=1 is neither point of Maxima nor point of Minima ∴ 𝒙=𝟏 is Point of Inflexion.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo