Question 4 Find the slope of the tangent to the curve 𝑦=𝑥^3−3𝑥+2 at the point whose 𝑥−coordinate is 3 𝑦=𝑥^3−3𝑥+2 We know that slope of tangent =𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥=3𝑥^2−3 Since 𝑥−coordinate is 3 Putting 𝑥=3 in (1) 〖𝑑𝑦/𝑑𝑥│〗_(𝑥 = 3)=3(3)^2−3 =3 ×9−3 =27−3 =24 Hence slope of tangent is 24

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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