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Ex 6.3

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Ex 6.3, 7 - Find points at which tangent is parallel to x-axis

Ex 6.3,7 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,7 - Chapter 6 Class 12 Application of Derivatives - Part 3

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Transcript

Ex 6.3, 7 Find points at which the tangent to the curve 𝑦=𝑥^3−3𝑥^2−9𝑥+7 is parallel to the x-axisEquation of Curve is 𝑦=𝑥^3−3𝑥^2−9𝑥+7 Differentiating w.r.t. 𝑥 𝑑𝑦/𝑑𝑥=𝑑(𝑥^3− 3𝑥^2 − 9𝑥 + 7)/𝑑𝑥 𝑑𝑦/𝑑𝑥=3𝑥^2−6𝑥−9+0 𝑑𝑦/𝑑𝑥=3𝑥^2−6𝑥−9 𝑑𝑦/𝑑𝑥=3(𝑥^2−2𝑥−3) 𝑑𝑦/𝑑𝑥=3(𝑥^2−3𝑥+𝑥−3) 𝑑𝑦/𝑑𝑥=3(𝑥(𝑥−3)+1(𝑥−3)) 𝑑𝑦/𝑑𝑥=3(𝑥+1)(𝑥−3) Given tangent to the curve is parallel to the 𝑥−𝑎𝑥𝑖𝑠 i.e. the Slope of tangent = Slope of 𝑥−𝑎𝑥𝑖𝑠 𝒅𝒚/𝒅𝒙=𝟎 3(𝑥+1)(𝑥−3)=0 (𝑥+1)(𝑥−3)=0 Thus 𝑥=−1 & 𝑥=3 When 𝒙=−𝟏 𝑦=𝑥^3−3𝑥^2−9𝑥+7 =(−1)^3−3(−1)^2−9(−1)+7 =−1−3+9+7 =12 Point is (−1 , 12) When 𝒙=𝟑 𝑦=𝑥^3−3𝑥^2−9𝑥+7 =(3)^3−3(3)^2−9(3)+7 =27−27−27+7 =− 20 Point is (3 , −20) Hence , the tangent to the Curve is parallel to the 𝑥−𝑎𝑥𝑖𝑠 at (−𝟏 , 𝟏𝟐) & (𝟑 , −𝟐𝟎)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.