Ex 6.5,10 - Chapter 6 Class 12 Application of Derivatives

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Ex 6.5,10 Find the maximum value of 2 3 24 + 107 in the interval [1, 3]. Find the maximum value of the same function in [ 3, 1]. Let f( )=2 ^3 24 +107 Step 1: Finding f ( ) f ( )= (2 ^3 24 +107)/ = 2 3 ^2 24 = 6 ^2 24 = 6 ( ^2 4" " ) Step 2: Putting f( )=0 6 ( ^2 4" " )=0 ^2 4" = 0 " ^2=4 = 4 = 2 Thus, =2 , 2 Step 3: Since x is in interval [1 , 3] = 2 is only Critical point Also, since given the interval = [1 , 3] We calculate f(x) at = 1 , 2 & 3 Hence maximum value of f( )= at = 3 in the interval [1 , 3] For the interval [ , ] = 2 is only Critical Point Also, since given the interval = [ 3, 1] We calculate f(x) at = 1 , 2 & 3 Hence maximum value of f( )= at = 2 in the interval [ 3, 1]