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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

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 Ex 6.5, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (ii) f (π‘₯) = sin⁑π‘₯ + cos⁑π‘₯ , π‘₯ ∈ [0, πœ‹ ] Finding f’(𝒙) f’(π‘₯)=𝑑(𝑠𝑖𝑛π‘₯ + π‘π‘œπ‘ π‘₯)/𝑑π‘₯ f’(π‘₯)=cos⁑〖π‘₯ βˆ’sin⁑π‘₯ γ€— Putting f’(𝒙) cos⁑〖π‘₯ βˆ’sin⁑π‘₯ γ€—= 0 cos⁑〖π‘₯=sin⁑π‘₯ γ€— 1 = sin⁑π‘₯/(cos⁑ π‘₯) 1 = tan π‘₯ tan π‘₯ = 1 We know that know tan ΞΈ = 1 at ΞΈ = πœ‹/4 ∴ π‘₯ = πœ‹/4 Since given interval π‘₯ ∈ [0 , πœ‹] Hence calculating f(π‘₯) at π‘₯=0 , πœ‹/4 ,πœ‹ Absolute Maximum value of f(π‘₯) is √𝟐 at 𝒙 = 𝝅/πŸ’ & Absolute Minimum value of f(π‘₯) is –1 at 𝒙 = Ο€ Ex 6.5, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (iii) f(π‘₯) = 4π‘₯ – 1/2 π‘₯2 , π‘₯ ∈ [βˆ’2, 9/2] f(π‘₯) = 4π‘₯ – 1/2 π‘₯2 Finding f’(𝒙) f’(π‘₯)=𝑑(4π‘₯ βˆ’ 1/2 π‘₯^2 )/𝑑π‘₯ = 4 – 1/2 Γ—2π‘₯ = 4 – π‘₯ Putting f’(𝒙)=𝟎 4 – π‘₯=0 π‘₯=4 ∴ π‘₯=4 is only critical point Since given interval π‘₯ ∈ [βˆ’2 , 9/2] Hence , calculating f(π‘₯) at π‘₯ = – 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 Ex 6.5, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (iv) f (π‘₯) = (π‘₯ βˆ’1)2 + 3, π‘₯ ∈ [βˆ’3,1] f (π‘₯) = (π‘₯ βˆ’1)2 + 3 Finding f’(𝒙) f’(x) = 𝑑((π‘₯ βˆ’ 1)^2+3)/𝑑π‘₯ = 2(π‘₯βˆ’1) Putting f’(𝒙)=𝟎 2(π‘₯βˆ’1)=0 π‘₯βˆ’1=0 π‘₯=1 Since given interval π‘₯ ∈ [βˆ’3 , 1] Hence , calculating f(π‘₯) at π‘₯ = – 3 , 1 Absolute Minimum value of f(x) is 3 at 𝒙 = 1 & Absolute Maximum value of f(x) is 19 at 𝒙 = – 3

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.