6.64. Expand the function $f(x)=x^{3}-4 x^{2}+5 x-2$ using the Taylor formula in powers of $(x+1)$, i.e., with the center of expansion at the point $x_{0}=-1$.
Solution.

$$
f(-1)=-12, f^{\prime}(x)=3 x^{2}-8 x+5, f^{\prime}(-1)=16, f^{\prime \prime}(x)=6 x-8
$$

$f^{\prime \prime}(-1)=-14, f^{\prime \prime \prime}(x)=6, f^{\prime \prime \prime}(-1)=6, f^{(n)}(x)=0$, if $n>3$.

## We have

$$
\begin{aligned}
f(x)= & x^{3}-4 x^{2}+5 x-2=f(-1)+\frac{f^{\prime}(-1)}{1!}(x+1)+\frac{f^{\prime \prime}(-1)}{2!}(x+1)^{2}+ \\
& +\frac{f^{\prime \prime \prime}(-1)}{3!}(x+1)^{3}+\ldots=-12+\frac{16}{1}(x+1)+\frac{-14}{2}(x+1)^{2}+ \\
& +\frac{6}{6}(x+1)^{3}+0=-12+16(x+1)-7(x+1)^{2}+(x+1)^{3}
\end{aligned}
$$