Check sibling questions

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

Introducing your new favourite teacher - Teachoo Black, at only β‚Ή83 per month


Transcript

Ex 6.3, 7 Find points at which the tangent to the curve 𝑦=π‘₯^3βˆ’3π‘₯^2βˆ’9π‘₯+7 is parallel to the x-axisEquation of Curve is 𝑦=π‘₯^3βˆ’3π‘₯^2βˆ’9π‘₯+7 Differentiating w.r.t. π‘₯ 𝑑𝑦/𝑑π‘₯=𝑑(π‘₯^3βˆ’ 3π‘₯^2 βˆ’ 9π‘₯ + 7)/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=3π‘₯^2βˆ’6π‘₯βˆ’9+0 𝑑𝑦/𝑑π‘₯=3π‘₯^2βˆ’6π‘₯βˆ’9 𝑑𝑦/𝑑π‘₯=3(π‘₯^2βˆ’2π‘₯βˆ’3) 𝑑𝑦/𝑑π‘₯=3(π‘₯^2βˆ’3π‘₯+π‘₯βˆ’3) 𝑑𝑦/𝑑π‘₯=3(π‘₯(π‘₯βˆ’3)+1(π‘₯βˆ’3)) 𝑑𝑦/𝑑π‘₯=3(π‘₯+1)(π‘₯βˆ’3) Given tangent to the curve is parallel to the π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 i.e. the Slope of tangent = Slope of π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 π’…π’š/𝒅𝒙=𝟎 3(π‘₯+1)(π‘₯βˆ’3)=0 (π‘₯+1)(π‘₯βˆ’3)=0 Thus π‘₯=βˆ’1 & π‘₯=3 When 𝒙=βˆ’πŸ 𝑦=π‘₯^3βˆ’3π‘₯^2βˆ’9π‘₯+7 =(βˆ’1)^3βˆ’3(βˆ’1)^2βˆ’9(βˆ’1)+7 =βˆ’1βˆ’3+9+7 =12 Point is (βˆ’1 , 12) When 𝒙=πŸ‘ 𝑦=π‘₯^3βˆ’3π‘₯^2βˆ’9π‘₯+7 =(3)^3βˆ’3(3)^2βˆ’9(3)+7 =27βˆ’27βˆ’27+7 =βˆ’ 20 Point is (3 , βˆ’20) Hence , the tangent to the Curve is parallel to the π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 at (βˆ’πŸ , 𝟏𝟐) & (πŸ‘ , βˆ’πŸπŸŽ)

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

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.