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

Last updated at Dec. 16, 2024 by Teachoo

   

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Ex 6.3,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (iii) 𝑓 (𝑥) = –(𝑥 – 1)^2+10 f(𝑥)=−(𝑥−1)^2+10 Finding f’ (x) Diff w.r.t 𝑥 f’(𝑥)=𝑑(−(𝑥−1)^2+10)/𝑑𝑥 f’(𝑥) = –2(𝑥−1)(𝑑(𝑥−1)/𝑑𝑥)+0 f’(𝑥) = –2(𝑥−1)(1−0) + 0 f’(𝑥)=−2(𝑥−1) Putting f’(𝒙)=𝟎 –2(𝑥−1)=0 (𝑥−1)=0 𝑥=1 Hence, 𝑥=1 is point of Maxima & No point of Minima Thus, f(𝑥) has maximum value at 𝑥=1 Putting x=1 in f(x) f(𝑥)=−(𝑥−1)^2+10 f(1)=−(1−1)^2+10 = 0 + 10 = 10 Maximum value of f(𝑥) is 10 There is no minimum value of f(𝒙) Ex 6.3,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (iii) 𝑓 (𝑥) = – (𝑥 – 1)^2+10 f(𝑥)=−(𝑥−1)^2+10 Finding f’(x) Diff w.r.t 𝑥 f’(𝑥)=𝑑(−(𝑥−1)^2+10)/𝑑𝑥 f’(𝑥) = –2(𝑥−1)(𝑑(𝑥−1)/𝑑𝑥)+0 f’(𝑥) = –2(𝑥−1)(1−0) + 0 f’(𝑥)=−2(𝑥−1) Putting f’(𝒙)=𝟎 –2(𝑥−1)=0 (𝑥−1)=0 𝑥=1 Finding f’’(𝒙) f’(𝑥)=−2(𝑥−1) Again diff w.r.t 𝑥 f’’(𝑥)=𝑑(−2(𝑥 − 1))/𝑑𝑥 =−2 𝑑(𝑥 − 1)/𝑑𝑥 =−2(1−0) =−2 Since f’’(𝑥) < 0 for 𝑥=1 Hence, f(𝑥) has Maximum value at 𝑥=1 Finding maximum value of f(𝒙) f(𝑥)=−(𝑥−1)^2+10 Putting 𝑥=1 f(𝑥) =−(1−1)^2+10 = 0 + 10 = 10 Maximum value of f(𝒙) is 10 There is no minimum value of f(𝒙)
  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

    3. Ex 6.3

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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