Ex 6.3, 7 - Find points at which tangent is parallel to x-axis - Finding point when tangent is parallel/ perpendicular

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Ex 6.3,7 Find points at which the tangent to the curve ๐‘ฆ=๐‘ฅ^3โˆ’3๐‘ฅ^2โˆ’9๐‘ฅ+7 is parallel to the x-axis Equation of Curve is ๐‘ฆ=๐‘ฅ^3โˆ’3๐‘ฅ^2โˆ’9๐‘ฅ+7 Differentiating w.r.t. ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^3โˆ’ 3๐‘ฅ^2 โˆ’ 9๐‘ฅ + 7)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=3๐‘ฅ^2โˆ’6๐‘ฅโˆ’9+0 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=3๐‘ฅ^2โˆ’6๐‘ฅโˆ’9 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=3(๐‘ฅ^2โˆ’2๐‘ฅโˆ’3) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=3(๐‘ฅ^2โˆ’3๐‘ฅ+๐‘ฅโˆ’3) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=3(๐‘ฅ(๐‘ฅโˆ’3)+1(๐‘ฅโˆ’3)) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=3(๐‘ฅ+1)(๐‘ฅโˆ’3) Given tangent to the curve is parallel to the ๐‘ฅโˆ’๐‘Ž๐‘ฅ๐‘–๐‘  i.e. the Slope of tangent = Slope of ๐‘ฅโˆ’๐‘Ž๐‘ฅ๐‘–๐‘  ๐’…๐’š/๐’…๐’™=๐ŸŽ 3(๐‘ฅ+1)(๐‘ฅโˆ’3)=0 (๐‘ฅ+1)(๐‘ฅโˆ’3)=0 Thus ๐‘ฅ=โˆ’1 & ๐‘ฅ=3 When ๐’™=โˆ’๐Ÿ ๐‘ฆ=๐‘ฅ^3โˆ’3๐‘ฅ^2โˆ’9๐‘ฅ+7 =(โˆ’1)^3โˆ’3(โˆ’1)^2โˆ’9(โˆ’1)+7 =โˆ’1โˆ’3+9+7 =12 Point is (โˆ’1 , 12) When ๐’™=๐Ÿ‘ ๐‘ฆ=๐‘ฅ^3โˆ’3๐‘ฅ^2โˆ’9๐‘ฅ+7 =(3)^3โˆ’3(3)^2โˆ’9(3)+7 =27โˆ’27โˆ’27+7 =โˆ’ 20 Point is (3 , โˆ’20) Hence , the tangent to the Curve is parallel to the ๐‘ฅโˆ’๐‘Ž๐‘ฅ๐‘–๐‘  at (โˆ’๐Ÿ , ๐Ÿ๐Ÿ) & (๐Ÿ‘ , โˆ’๐Ÿ๐ŸŽ)

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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.