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Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 16

Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 17
Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 18
Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 19
Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 20
Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 21


Transcript

Ex 6.5,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (iii) 𝑓 (π‘₯) = –(π‘₯ – 1)^2+10 f(π‘₯)=βˆ’(π‘₯βˆ’1)^2+10 Finding f’ (x) Diff w.r.t π‘₯ f’(π‘₯)=𝑑(βˆ’(π‘₯βˆ’1)^2+10)/𝑑π‘₯ f’(π‘₯) = –2(π‘₯βˆ’1)(𝑑(π‘₯βˆ’1)/𝑑π‘₯)+0 f’(π‘₯) = –2(π‘₯βˆ’1)(1βˆ’0) + 0 f’(π‘₯)=βˆ’2(π‘₯βˆ’1) Putting f’(𝒙)=𝟎 –2(π‘₯βˆ’1)=0 (π‘₯βˆ’1)=0 π‘₯=1 Hence, π‘₯=1 is point of Maxima & No point of Minima Thus, f(π‘₯) has maximum value at π‘₯=1 Putting x=1 in f(x) f(π‘₯)=βˆ’(π‘₯βˆ’1)^2+10 f(1)=βˆ’(1βˆ’1)^2+10 = 0 + 10 = 10 Maximum value of f(π‘₯) is 10 There is no minimum value of f(𝒙) Ex 6.5,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (iii) 𝑓 (π‘₯) = – (π‘₯ – 1)^2+10 f(π‘₯)=βˆ’(π‘₯βˆ’1)^2+10 Finding f’(x) Diff w.r.t π‘₯ f’(π‘₯)=𝑑(βˆ’(π‘₯βˆ’1)^2+10)/𝑑π‘₯ f’(π‘₯) = –2(π‘₯βˆ’1)(𝑑(π‘₯βˆ’1)/𝑑π‘₯)+0 f’(π‘₯) = –2(π‘₯βˆ’1)(1βˆ’0) + 0 f’(π‘₯)=βˆ’2(π‘₯βˆ’1) Putting f’(𝒙)=𝟎 –2(π‘₯βˆ’1)=0 (π‘₯βˆ’1)=0 π‘₯=1 Finding f’’(𝒙) f’(π‘₯)=βˆ’2(π‘₯βˆ’1) Again diff w.r.t π‘₯ f’’(π‘₯)=𝑑(βˆ’2(π‘₯ βˆ’ 1))/𝑑π‘₯ =βˆ’2 𝑑(π‘₯ βˆ’ 1)/𝑑π‘₯ =βˆ’2(1βˆ’0) =βˆ’2 Since f’’(π‘₯) < 0 for π‘₯=1 Hence, f(π‘₯) has Maximum value at π‘₯=1 Finding maximum value of f(𝒙) f(π‘₯)=βˆ’(π‘₯βˆ’1)^2+10 Putting π‘₯=1 f(π‘₯) =βˆ’(1βˆ’1)^2+10 = 0 + 10 = 10 Maximum value of f(𝒙) is 10 There is no minimum value of f(𝒙)

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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.