Example 20 - Chapter 6 Class 12 Application of Derivatives

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Example 20 Find local maximum and local minimum values of the function f given by f (𝑥)=3𝑥4 + 4𝑥3 – 12𝑥2 + 12f (𝑥)=3𝑥4 + 4𝑥3 – 12𝑥2 + 12 Finding f’ (𝒙) f’ (𝑥)=𝑑(3𝑥4 + 4𝑥3 – 12𝑥2 + 12)/𝑑𝑥 f’ (𝑥)=12𝑥^3+12𝑥^2 – 24𝑥 "+ 0" f’ (𝑥)=12(𝑥^3+𝑥^2−2𝑥) f’ (𝑥)=12𝑥(𝑥^2+𝑥−2) f’ (𝑥)=12𝑥 (𝑥^2+2𝑥−𝑥−2) f’ (𝑥)=12𝑥 (𝑥(𝑥+2)−1(𝑥+2)) f’ (𝑥)=𝟏𝟐𝒙 (𝒙−𝟏)(𝒙+𝟐) Putting f’ (𝒙)=𝟎 12𝑥 (𝑥−1)(𝑥+2)=0 𝑥 (𝑥−1)(𝑥+2)=0 So, 𝒙=𝟎,𝑥=𝟏,& 𝑥=−𝟏 Finding f’’(𝒙) f ’(𝑥)=12(𝑥^3+𝑥^2−2𝑥) f ’’(𝑥)=12𝑑(𝑥^3 + 𝑥^2 − 2𝑥)/𝑑𝑥 f ’’(𝑥)=𝟏𝟐(𝟑𝒙^𝟐+𝟐𝒙−𝟐) At 𝒙=𝟎 f ’’(0)=12(3(0)^2+2(0)−2)= 32 (0+0 −2)= – 64 < 0 Since f’’(𝑥)<0 at 𝑥=0 ∴ 𝑥 = 0 is point of local maxima Thus, f(𝑥) is maximum at 𝑥=0 At 𝒙=𝟏 f’’(1)=12(3(1)^2+2(1)−2)= 12 (3+2−2) = 36 > 0 Since f’’(𝑥)>0 at 𝑥=1 ∴ 𝑥 = 1 is point of local minima Thus, f(𝑥) is minimum at 𝑥=1 At 𝒙=−𝟐 f’’(−2)=12(3(−2)^2+2(−2)−2)= 12 (12−4−2)= 72 > 0 Since f’’(𝑥)>0 at 𝑥=−2 ∴ 𝑥 = −2 is point of local minima Thus, f(𝑥) is minimum at 𝑥=−2 Finding local minimum and maximum value f’ (𝑥)=𝟏𝟐𝒙 (𝒙−𝟏)(𝒙+𝟐) Local maximum value of f (𝑥) at 𝑥=0 f (0)=3(0)4 + 4(0)3 – 12(0)2 + 12 = 0 + 0 – 0 + 12 = 12 Local minimum value of f (𝑥) at 𝑥=1 f (1)=3(1)4 + 4(1)3 – 12(1)2 + 12 = 3 + 4 – 12 + 12 = 7 Local Minimum value of f (𝑥) at 𝑥=−2 f (−2)=3(−2)4 + 4(−2)3 – 12(−2)2 + 12 = 48 – 32 – 48 + 12 = – 20

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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