Ex 6.3, 5 (iii) - Chapter 6 Class 12 Application of Derivatives

Last updated at August 25, 2026 by

  Ex 6.3, 5 (iii) - For f(x) = 4x - 1/2 x^2, x ∈ [-2, 9/2], find absolut - Ex 6.3

part 2 - Ex 6.3, 5 (iii) - Ex 6.3 - Serial order wise - Chapter 6 Class 12 Application of Derivatives
part 3 - Ex 6.3, 5 (iii) - Ex 6.3 - Serial order wise - Chapter 6 Class 12 Application of Derivatives

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Transcript

Ex 6.3, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (iii) f(𝑥) = 4𝑥 – 1/2 𝑥2 , 𝑥 ∈ [−2, 9/2] Putting f’(𝒙)=𝟎 4 – 𝑥=0 𝑥=4 ∴ 𝑥=4 is only critical point Since given interval 𝑥 ∈ [−2 , 9/2] Hence , calculating f(𝑥) at 𝑥 = – 2 , 4 , 9/2 Absolute Maximum value of f(x) is 8 at 𝒙 = 4 & Absolute Minimum value of f(x) is –10 at 𝒙 = –2

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