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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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