Ex 6.5,7 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 15, 2021 by Teachoo

Ex 6.5, 7 - Find both max, min value of 3x4 - 8x3 + 12x2

Ex 6.5,7 - Chapter 6 Class 12 Application of Derivatives - Part 2

Ex 6.5,7 - Chapter 6 Class 12 Application of Derivatives - Part 3

Ex 6.5,7 - Chapter 6 Class 12 Application of Derivatives - Part 4

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Transcript

Ex 6.5, 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 Square of number can not be negative 𝑥^2+2=0 𝑥^2=−2 𝑥−2=0 𝑥=2 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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