Set [tex]f(x)=2x^3-3x^2+2[/tex]. Find the tangent line [tex]\ell_1(x)[/tex] to [tex]f(x)[/tex] at the point when [tex]x=x_1[/tex]:
[tex]f'(x)=6x^2-6x\implies f'(x_1)=12[/tex] (slope of [tex]\ell_1[/tex])
[tex]\implies\ell_1(x)=12(x-x_1)+f(x_1)=12(x+1)-3=12x+9[/tex]
Set [tex]x_2=-\dfrac9{12}[/tex], the root of [tex]\ell_1(x)[/tex]. The tangent line [tex]\ell_2(x)[/tex] to [tex]f(x)[/tex] at [tex]x=x_2[/tex] has slope and thus equation
[tex]f'(x_2)=\dfrac{63}8\implies\ell_2(x)=\dfrac{63}8\left(x+\dfrac9{12}\right)-\dfrac{17}{32}=7x+\dfrac{151}{32}[/tex]
which has its root at [tex]x_3=-\dfrac{151}{224}\approx-0.6741[/tex].
(The actual value of this root is about -0.6777)
