1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
  3. Ex 6.3
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Ex 6.3,10 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 16, 2024 by Teachoo

 

Transcript

Ex 6.3, 10 Find the maximum value of 2𝑥3 – 24𝑥 + 107 in the interval [1, 3]. Find the maximum value of the same function in [–3, –1]. Let f(𝑥)=2𝑥^3−24𝑥+107 Finding f’(𝑥) f’(𝑥)=𝑑(2𝑥^3 − 24𝑥 + 107)/𝑑𝑥 = 2 × 3𝑥^2−24 = 6𝑥^2−24 = 6 (𝑥^2−4" " ) Putting f(𝑥)=0 6 (𝑥^2−4" " )=0 𝑥^2−4" = 0 " 𝑥^2=4 𝑥=±√4 𝑥=±2 Thus, 𝑥=2 , – 2 Since x is in interval [1 , 3] 𝑥 = 2 is only Critical point Also, since given the interval 𝑥= ∈ [1 , 3] We calculate f(x) at 𝑥= 1 , 2 & 3 Hence maximum value of f(𝒙)=𝟖𝟗 at 𝒙 = 3 in the interval [1 , 3] For the interval [−𝟑 , −𝟏] 𝑥 = –2 is only Critical Point Also, since given the interval 𝑥= ∈ [−3,−1] We calculate f(x) at 𝑥= –1 , –2 & –3 Hence maximum value of f(𝒙)=𝟏𝟑𝟗 at 𝒙 = –2 in the interval [−3,−1]
  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

    3. Ex 6.3

    Examples →
Examples →
Chapter 6 Class 12 Application of Derivatives
Serial order wise

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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