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Ex 6.5, 5 - Find absolute maximum and min value of - Class 12

Ex 6.5,5 - Chapter 6 Class 12 Application of Derivatives - Part 2



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Ex 6.5, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (i) f (π‘₯) = π‘₯3, π‘₯∈[– 2, 2]Finding f’(𝒙) f’(π‘₯)=𝑑(π‘₯^3 )/𝑑π‘₯ f’(π‘₯)=3π‘₯^2 Putting f’(𝒙)=𝟎 3π‘₯^2=0 π‘₯^2=0 π‘₯=0 So, π‘₯=0 is critical point Since given interval is π‘₯ ∈ [βˆ’2 , 2] Hence calculating f(π‘₯) at π‘₯=βˆ’2 , 0 , 2 Absolute Maximum value of f(x) is 8 at 𝒙 = 2 & Absolute Minimum value of f(x) is –8 at 𝒙 = –2

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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.