Ex 6.3,7 - Chapter 6 Class 12 Application of Derivatives

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Ex 6.3, 7 Find both the maximum value and the minimum value of 3𝑥4 – 8𝑥3 + 12𝑥2 – 48𝑥 + 25 on the interval [0, 3].Let f(x) = 3𝑥4 – 8𝑥3 + 12𝑥2 – 48𝑥 + 25, where 𝑥 ∈ [0, 3] Finding f’(𝒙) f’(𝑥)=𝑑(3𝑥^4 − 8𝑥^3 + 12𝑥^2 − 48𝑥 + 25)/𝑑𝑥 f’(𝑥)=3 ×4𝑥^3−8 ×3𝑥^2+12 ×2𝑥−48+0 f’(𝑥)=12𝑥^3−24𝑥^2+24𝑥−48 f’(𝑥)=12(𝑥^3−2𝑥^2+2𝑥−4) Putting f’(𝒙)=𝟎 12(𝑥^3−2𝑥^2+2𝑥−4)=0 𝑥^3−2𝑥^2+2𝑥−4=0 𝑥^2 (𝑥−2)+2(𝑥−2)=0 (𝑥^2+2)(𝑥−2)=0 Since 𝑥^2=−2 is not possible Thus 𝑥=2 is only critical point Since are given interval 𝑥 ∈ [0 , 3] Hence , calculating f(𝑥) at 𝑥 = 0 , 2 & 3 Hence, Minimum value of f(𝑥) is –39 at 𝒙 = 2 Maximum value of f(𝑥) is 25 at 𝒙 = 0

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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