Ex 6.5, 5 - Find absolute maximum and min value of - Class 12 - Ex 6.5

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Ex 6.5,5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (i) f ( ) = 3, [ 2, 2] Step 1: Finding f f = 3 f =3 2 Step 2: Putting f =0 3 2 =0 2 =0 =0 So, =0 is critical point Step 3: Since given interval 2 , 2 Hence calculating f at = 2 , 0 , 2 Step 4: Absolute Maximum value of f(x) is 8 at = 2 & Absolute Minimum value of f(x) is 8 at = 2 Ex 6.5,5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (ii) f ( ) = sin + cos , [0, ] Step 1: Finding f f = + f = cos sin Step 2: Putting f cos sin = 0 cos = sin 1 = sin cos 1 = tan tan = 1 We know that know tan = 1 at = 4 = 4 Step 3: Since given interval 0 , Hence calculating f at =0 , 4 , Absolute Maximum value of f is at = & Absolute Minimum value of f is 1 at = Ex 6.5,5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (iii) f( ) = 4 1 2 2 , 2, 9 2 f( ) = 4 1 2 2 Step 1: Finding f f = 4 1 2 2 = 4 1 2 2 = 4 Step 2: Putting f =0 4 =0 =4 =4 is only critical point Step 3: Since given interval 2 , 9 2 Hence , calculating f at = 2 , 4 , 9 2 Absolute Maximum value of f(x) is 8 at = 4 & Absolute Minimum value of f(x) is 10 at = 2 Ex 6.5,5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (iv) f ( ) = ( 1)2 + 3, [ 3,1] f ( ) = ( 1)2 + 3 Step 1: Finding f f = 1 2 +3 = 2 1 Step 2: Putting f =0 2 1 =0 1=0 =1 Step 3: Since given interval 3 , 1 Hence , calculating f at = 3 , 1 Absolute Minimum value of f(x) is 3 at = 1 & Absolute Maximum value of f(x) is 19 at = 3

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.