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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


Transcript

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 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.