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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

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.