Ex 6.5, 10 - Find max value of 2x3 - 24 x + 107 in [1, 3] - Ex 6.5

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Ex 6.5,10 Find the maximum value of 2๐‘ฅ3 โ€“ 24๐‘ฅ + 107 in the interval [1, 3]. Find the maximum value of the same function in [โ€“3, โ€“1]. Let f(๐‘ฅ)=2๐‘ฅ^3โˆ’24๐‘ฅ+107 Step 1: Finding fโ€™(๐‘ฅ) fโ€™(๐‘ฅ)=๐‘‘(2๐‘ฅ^3โˆ’24๐‘ฅ+107)/๐‘‘๐‘ฅ = 2 ร— 3๐‘ฅ^2โˆ’24 = 6๐‘ฅ^2โˆ’24 = 6 (๐‘ฅ^2โˆ’4" " ) Step 2: Putting f(๐‘ฅ)=0 6 (๐‘ฅ^2โˆ’4" " )=0 ๐‘ฅ^2โˆ’4" = 0 " ๐‘ฅ^2=4 ๐‘ฅ=ยฑโˆš4 ๐‘ฅ=ยฑ2 Thus, ๐‘ฅ=2 , โ€“ 2 Step 3: Since x is in interval [1 , 3] ๐‘ฅ = 2 is only Critical point Also, since given the interval ๐‘ฅ= โˆˆ [1 , 3] We calculate f(x) at ๐‘ฅ= 1 , 2 & 3 Hence maximum value of f(๐’™)=๐Ÿ–๐Ÿ— at ๐’™ = 3 in the interval [1 , 3] For the interval [โˆ’๐Ÿ‘ , โˆ’๐Ÿ] ๐‘ฅ = โ€“2 is only Critical Point Also, since given the interval ๐‘ฅ= โˆˆ [โˆ’3,โˆ’1] We calculate f(x) at ๐‘ฅ= โ€“1 , โ€“2 & โ€“3 Hence maximum value of f(๐’™)=๐Ÿ๐Ÿ‘๐Ÿ— at ๐’™ = โ€“2 in the interval [โˆ’3,โˆ’1]

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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.