$(1)$ Due date $12/9/248$:$00$ AM

DF w$/$o expl

$26.$ CalC$12$j$14$a $10$ pts possible

Find the degree three Taylor polynomial $T_3$ entered at $x=0$ for $f$ when

$f(x)=ln(3-2x)$

$T_3(x)=ln3-\frac{2}{3}x+\frac{2}{9}{x}^2-\frac{8}{81}{x}^3$

$T_3(x)=\frac{2}{3}x-\frac{2}{9}{x}^2-\frac{8}{81}{x}^3$

$T_3(x)=ln3+\frac{2}{3}x-\frac{2}{9}{x}^2+\frac{4}{81}{x}^3$

$T_3(x)=\frac{2}{3}x-\frac{2}{9}{x}^2+\frac{8}{81}{x}^3$

$T_3(x)=\frac{2}{3}x+\frac{2}{9}{x}^2+\frac{8}{81}{x}^3$

$T_3(x)=ln3-\frac{2}{3}x-\frac{2}{9}{x}^2-\frac{8}{81}{x}^3$

Find the degree three Taylor polynomial $T_3$ centered at $x=0$ for $f$ when

$f(x)=ln(3-2x)$

$\backslash$times $1.{?}^{T_3(x)}=ln3-\frac{2}{3}x+\frac{2}{9}{x}^2-\frac{8}{81}{x}^3$

$2.T_3(x)=\frac{2}{3}x-\frac{2}{9}{x}^2-\frac{8}{81}{x}^3$

$3.T_3(x)=ln3+\frac{2}{3}x-\frac{2}{9}{x}^2+\frac{4}{81}{x}^3$

$4.T_3(x)=\frac{2}{3}x-\frac{2}{9}{x}^2+\frac{8}{81}{x}^3$

$5.T_3(x)=\frac{2}{3}x+\frac{2}{9}{x}^2+\frac{8}{81}{x}^3$

$6.T_3(x)=ln3-\frac{2}{3}x-\frac{2}{9}{x}^2-\frac{8}{81}{x}^3$