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Ex 6.5, 7 - Find both max, min value of 3x4 - 8x3 + 12x2

Ex 6.5,7 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5,7 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.5,7 - Chapter 6 Class 12 Application of Derivatives - Part 4


Transcript

Ex 6.5, 7 Find both the maximum value and the minimum value of 3π‘₯4 – 8π‘₯3 + 12π‘₯2 – 48π‘₯ + 25 on the interval [0, 3].Let f(x) = 3π‘₯4 – 8π‘₯3 + 12π‘₯2 – 48π‘₯ + 25, where π‘₯ ∈ [0, 3] Finding f’(𝒙) f’(π‘₯)=𝑑(3π‘₯^4 βˆ’ 8π‘₯^3 + 12π‘₯^2 βˆ’ 48π‘₯ + 25)/𝑑π‘₯ f’(π‘₯)=3 Γ—4π‘₯^3βˆ’8 Γ—3π‘₯^2+12 Γ—2π‘₯βˆ’48+0 f’(π‘₯)=12π‘₯^3βˆ’24π‘₯^2+24π‘₯βˆ’48 f’(π‘₯)=12(π‘₯^3βˆ’2π‘₯^2+2π‘₯βˆ’4) Putting f’(𝒙)=𝟎 12(π‘₯^3βˆ’2π‘₯^2+2π‘₯βˆ’4)=0 π‘₯^3βˆ’2π‘₯^2+2π‘₯βˆ’4=0 π‘₯^2 (π‘₯βˆ’2)+2(π‘₯βˆ’2)=0 (π‘₯^2+2)(π‘₯βˆ’2)=0 Since π‘₯^2=βˆ’2 is not possible Thus π‘₯=2 is only critical point Square of number can not be negative π‘₯^2+2=0 π‘₯^2=βˆ’2 π‘₯βˆ’2=0 π‘₯=2 Since are given interval π‘₯ ∈ [0 , 3] Hence , calculating f(π‘₯) at π‘₯ = 0 , 2 & 3 Hence, Minimum value of f(π‘₯) is –39 at 𝒙 = 2 Maximum value of f(π‘₯) is 25 at 𝒙 = 0

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.