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Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 9

Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 10
Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 11 Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 12 Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 13 Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 14 Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 15

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Ex 6.5, 1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (ii) f (π‘₯) = 9π‘₯2+12π‘₯+2Finding f’(x) f (π‘₯)=9π‘₯2+12π‘₯+2 Diff. w.r.t π‘₯ f’(π‘₯)=18π‘₯+12 f’(π‘₯)=6(3π‘₯+2) Putting f’(𝒙)=𝟎 6(3π‘₯+2)=0 3π‘₯+2=0 3π‘₯=βˆ’2 π‘₯=(βˆ’2)/( 3) Hence π‘₯=(βˆ’2)/3 is point of minima of f(π‘₯) Finding minimum value of f(π‘₯) at π‘₯=(βˆ’2)/3 f(π‘₯)=9π‘₯^2+12π‘₯+2 Putting π‘₯=(βˆ’2)/3 f(π‘₯)=9((βˆ’2)/3)^2+12((βˆ’2)/3)+2=9(4/3)βˆ’12(2/3)+2=βˆ’2 Thus, Minimum value of f(𝒙)=βˆ’πŸ There is no maximum value Ex 6.5, 1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (ii) f (π‘₯) = 9π‘₯2+12π‘₯+2Finding f’(𝒙) f (π‘₯)=9π‘₯2+12π‘₯+2 Diff. w.r.t π‘₯ f’(π‘₯)=𝑑(9π‘₯^2 + 12π‘₯ + 2)/𝑑π‘₯ f’(π‘₯)=18π‘₯+12 f’(π‘₯)=6(3π‘₯+2) Putting f’(𝒙)=𝟎 6(3π‘₯+2)=0 3π‘₯+2=0 3π‘₯=βˆ’2 π‘₯=(βˆ’2)/3 Finding f’’(𝒙) f’(π‘₯)= 6(3π‘₯+2) Again diff w.r.t π‘₯ f’’(π‘₯)=𝑑(6(3π‘₯+2))/𝑑π‘₯ f’’(π‘₯)=6 𝑑(3π‘₯ + 2)/𝑑π‘₯ f’’(π‘₯)=6(3+0) f’’(π‘₯)=6(3) f’’(π‘₯)=18 So, f’’((βˆ’2)/3)=18 Since f’’(π‘₯)>0 is for π‘₯=(βˆ’2)/3 π‘₯=(βˆ’2)/3 is point of local minima Finding minimum value Putting π‘₯=(βˆ’2)/3 in f(π‘₯) f (π‘₯)=9π‘₯2+12π‘₯+2 f ((βˆ’2)/3)=9((βˆ’2)/3)^2+12((βˆ’2)/3)+2 =9(4/9)+12((βˆ’2)/3)+2 =4βˆ’8+2 =βˆ’2 Hence, minimum value = –2 There is no maximum value

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.