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

Question

Elijah Abram · Accepted Answer

#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#

Paisley Johnson · Answer

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]

William Abdalla · Answer

undefined To find the equation of the tangent line of f(x)=(x-2)(x-2)(lnx-x) at x=3, we need to find the slope of the tangent line and a point on the line.

First, we find the derivative of f(x) using the product rule and the chain rule:

f'(x) = (x-2)(2lnx - 1) + (x-2)(1/x) + (lnx - x)(x-2)(1/x)

Next, we substitute x=3 into f'(x) to find the slope of the tangent line:

f'(3) = (3-2)(2ln3 - 1) + (3-2)(1/3) + (ln3 - 3)(3-2)(1/3)

Finally, we substitute the slope and the point (3, f(3)) into the point-slope form of a line to find the equation of the tangent line.