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Ex 6.3

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Ex 6.3, 16 - Show that tangents to y = 7x3 + 11 at x = 2, -2

Ex 6.3,16 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,16 - Chapter 6 Class 12 Application of Derivatives - Part 3


Transcript

Ex 6.3, 16 Show that the tangents to the curve 𝑦=7𝑥3+11 at the points where 𝑥=2 and 𝑥 =−2 are parallel.We know that 2 lines are parallel y Slope of 1st line = Slope of 2nd line 𝑚1=𝑚2 We know that Slope of tangent is 𝑑𝑦/𝑑𝑥 Given Curve is 𝑦=7𝑥^3+11 Differentiating w.r.t.𝑥 𝑑𝑦/𝑑𝑥=𝑑(7𝑥3 + 11)/𝑑𝑥 𝑑𝑦/𝑑𝑥=21𝑥^2 We need to show that tangent at 𝑥=2 & tangent at 𝑥=−2 are parallel i.e. we need to show (Slope of tangent at 𝑥=2) = (Slope of tangent at 𝑥=−2) Now, Slope of tangent = 𝑑𝑦/𝑑𝑥=21𝑥^2 Slope of tangent at 𝑥=2 〖𝑑𝑦/𝑑𝑥│〗_(𝑥 = 2)=21(2)^2=21 ×4=84 & Slope of tangent at 𝑥=−2 〖𝑑𝑦/𝑑𝑥│〗_(𝑥 =− 2)=21(−2)^2=21 ×4=84 Since, (Slope of tangent at 𝑥=2) = (Slope of tangent at 𝑥=−2) Thus, tangent at 𝑥=2 & tangent at 𝑥 are parallel Hence Proved

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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.