What is the equation of the tangent line of #f(x)=(x-2)(x-2)(lnx-x)# at #x=3#?

Question

1411 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-30T11:59:21

#y=(-2/3+2(In3-3))x-5In3+17#

#f(x)=(x-2)(x-2)(Inx-x)#
#f(x)=(x-2)^2(Inx-x)#
#f'(x)=(x-2)^2(1/x-1)+(Inx-x)times2(x-2)#
#f'(x)=(x-2)^2(1/x-1)+2(x-2)(Inx-x)#

At #x=3#,
#f'(3)=(3-2)^2(1/3-1)+2(3-2)(In3-3)#
#f'(3)=-2/3+2(In3-3)#

Equation of the tangent at #(3, In3-3)#

#y-(In3-3)=(-2/3+2(In3-3))(x-3)#

#y-In3+3=(-2/3+2(In3-3))x+2-6(In3-3)#

#y-In3+3=(-2/3+2(In3-3))+2-6In3+18#

#y-In3=(-2/3+2(In3-3))x-6In3+17#

#y=(-2/3+2(In3-3))x-5In3+17#

Answer 2 · 2018-07-30T12:26:58

Equation of tangent is #y = -4.47 x +11.5#

#f(x)= (x-2)(x-2)(ln x- x) ; x=3# or

#f(x)= (x-2)^2(ln x- x) #

#f(3)= (3-2)^2(ln 3- 3) ~~ -1.90# The point at which ,tangent

to be drawn is # (3, -1.9)#

#f(x)= (x-2)^2(ln x- x) #

#f'(x)= (x-2)^2(1/x- 1) + 2(x-2)(ln x -x)#

#f'(3)= (3-2)^2(1/3- 1) + 2(3-2)(ln 3 -3)~~ -4.47#

Slope of curve at # (3, -1.9)# is #m=-4.47#

Equation of tangent is # y-y_1= m(x-x_1)# , therefore

equation of tangent is # y-(-1.9)= -4.47(x-3)# or

#y+1.9= -4.47 x +13.4 or y = -4.47 x +11.5# [Ans]