Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12      1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Examples

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Example 40 Find absolute maximum and minimum values of a function f given by f (𝑥) = 〖12 𝑥〗^(4/3) – 〖 6𝑥〗^(1/3) , 𝑥 ∈ [ – 1, 1] f (𝑥) = 〖12 𝑥〗^(4/3) – 〖 6𝑥〗^(1/3) Finding f’(𝒙) f’(𝑥)=𝑑(12𝑥^(4/3) − 6𝑥^(1/3) )/𝑑𝑥 = 12 × 4/3 𝑥^(4/3 −1)−6 × 1/3 𝑥^(1/3 −1) = 4 × 4 𝑥^((4 − 3)/3) −2𝑥^((1 − 3)/3) = 16 𝑥^(1/3) −2𝑥^((−2)/3) = 16 𝑥^(1/3) − 2/𝑥^(2/3) = (16𝑥^(1/3) × 𝑥^(2/3) − 2)/𝑥^(2/3) = (16𝑥^(1/3 + 2/3) − 2)/𝑥^(2/3) = (16𝑥^(3/3) − 2)/𝑥^(2/3) = (16𝑥 − 2)/𝑥^(2/3) = 2(8𝑥 − 1)" " /𝑥^(2/3) Hence, f’(𝑥)=2(8𝑥 − 1)/𝑥^(2/3) Putting f’(𝒙)=𝟎 2(8𝑥 − 1)/𝑥^(2/3) =0 2(8𝑥−1)=0 ×𝑥^(2/3) 2(8𝑥−1)=0 8𝑥−1= 0 8𝑥=1 𝑥=1/8 Note that: Since f’(𝑥)=2(8𝑥 − 1)/𝑥^(2/3) f’(𝑥) is not defined at 𝑥= 0 𝑥=1/8 & 0 are critical points We are given interval [−1 , 1] Hence calculating f(𝑥) at 𝑥=0, 1/8, −1 , 1 𝑥 = 0 f(0)=12(0)^(4/3)−6(0)^(1/3) = 0 – 0 = 0 𝒙 = 𝟏/𝟖 f(1/8)=12(1/8)^(4/3)−6(1/8)^(1/3) = 12 × (1/2)^(3 × 4/3)−6(1/2)^(3 × 4/3) = 12(1/2)^4 – 6(1/2)^1 = (−𝟗)/𝟒 𝑥 = 1 f(1)=12(1)^(4/3)−6(1)^(1/3) = 12 – 6 = 6 Hence, Absolute maximum value of f(x) is 18 at 𝒙 = –1 & Absolute minimum value of f(x) is (−9)/4 at 𝒙 = 𝟏/𝟖

Examples 