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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 6.3, 5 (i) - Chapter 6 Class 12 Application of Derivatives
Last updated at May 29, 2023 by Teachoo
Ex 6.3
Ex 6.3, 1 (i)
Important
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
Ex 6.3, 2 (iii)
Ex 6.3, 2 (iv) Important
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Ex 6.3, 3 (i)
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Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important
Ex 6.3, 3 (v)
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important
Ex 6.3, 3 (viii)
Ex 6.3, 4 (i)
Ex 6.3, 4 (ii) Important
Ex 6.3, 4 (iii)
Ex 6.3, 5 (i) You are here
Ex 6.3, 5 (ii)
Ex 6.3, 5 (iii) Important
Ex 6.3, 5 (iv)
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8
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Ex 6.3, 26 Important
Ex 6.3, 27 (MCQ)
Ex 6.3,28 (MCQ) Important
Ex 6.3,29 (MCQ)
Transcript
Ex 6.3, 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₯β[β 2, 2]Finding fβ(π) fβ(π₯)=π(π₯^3 )/ππ₯ fβ(π₯)=3π₯^2 Putting fβ(π)=π 3π₯^2=0 π₯^2=0 π₯=0 So, π₯=0 is critical point Since given interval is π₯ β [β2 , 2] Hence calculating f(π₯) at π₯=β2 , 0 , 2 Absolute Maximum value of f(x) is 8 at π = 2 & Absolute Minimum value of f(x) is β8 at π = β2
