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Ex 6.3, 9 - Find point on y = x3 - 11x + 5 at which tangent

Ex 6.3,9 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,9 - Chapter 6 Class 12 Application of Derivatives - Part 3

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Transcript

Ex 6.3, 9 Find the point on the curve 𝑦=π‘₯^3βˆ’11π‘₯+5 at which the tangent is 𝑦=π‘₯ βˆ’11.Equation of Curve is 𝑦=π‘₯^3βˆ’11π‘₯+5 We know that Slope of tangent is 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=𝑑(π‘₯^3 βˆ’ 11π‘₯ + 5)/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=γ€–3π‘₯γ€—^2βˆ’11 Also, Given tangent is 𝑦=π‘₯βˆ’12 Comparing with 𝑦=π‘šπ‘₯+𝑐 , when m is the Slope Slope of tangent =1 From (1) and (2) 𝑑𝑦/𝑑π‘₯=1 3π‘₯^2βˆ’11=1 3π‘₯^2=1+11 3π‘₯^2=12 π‘₯^2=12/3 π‘₯^2=4 π‘₯=Β±2 When 𝒙=𝟐 𝑦=(2)^3βˆ’11(2)+5 𝑦=8βˆ’22+5 𝑦=βˆ’ 9 So, Point is (2 , βˆ’9) When 𝒙=βˆ’πŸ 𝑦=(βˆ’2)^3βˆ’11(βˆ’2)+5 𝑦=βˆ’ 8+22+5 𝑦=19 So, Point is (βˆ’2 , 19) Hence , Points (2 , βˆ’9) & (βˆ’2 , 19) But (–2, 19) does not satisfy line y = x – 11 As 19 β‰  –2 – 11 ∴ Only point is (2, –9)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.