Example 18 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 16, 2024 by

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Example 18 (Method 1) 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)=6𝑥2 –12𝑥+6+0 f ′(𝑥)=6(𝑥^2−2𝑥+1) Putting f ′(𝒙)= 0 6(𝑥^2−2𝑥+1)=0 𝑥^2−2𝑥+1=0 (𝑥)^2+(1)^2−2(𝑥)(1)=0 (𝑥−1)^2=0 So, 𝒙=𝟏 is only critical point Hence 𝒙=𝟏 is point of inflexion Example 18 (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)=6𝑥2 –12𝑥+6+0 f ′(𝑥)=6(𝑥^2−2𝑥+1) Putting f ′(𝒙)= 0 6(𝑥^2−2𝑥+1)=0 𝑥^2−2𝑥+1=0 𝑥^2+1^2−2(𝑥)(1)=0 (𝑥−1)^2=0 So, 𝒙=𝟏 is only critical point Finding f’’(𝒙) f’’(𝑥)=6 𝑑(𝑥^2 − 2𝑥 + 1)/𝑑𝑥 f’’(𝑥)=6(2𝑥−2+0) f’’(𝑥)=12(𝑥−1) Putting 𝒙=𝟏 f’’(1)=12(1−1) = 12 × 0 = 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

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