Question

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

Answer 1

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

Explanation:

`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

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

Explanation:

`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]