## Calculus (3rd Edition)

$$T_{2}(x) =\frac{1}{e}+\frac{1}{e}(x-1)+\frac{-1 / e}{2}(x-1)^{2}$$ $$T_{3}(x) =\frac{1}{e}+\frac{1}{e}(x-1)-\frac{1}{2 e}(x-1)^{2}-\frac{1}{6 e}(x-1)^{3}$$
Given $$f(x)=x^{2} e^{-x}, \quad a=1$$ Since \begin{aligned} f(x) &=x^{2} e^{-x} & & f(1) &=1 / e \\ f^{\prime}(x) &=\left(2 x-x^{2}\right) e^{-x} & & f^{\prime}(1) &=1 / e \\ f^{\prime \prime}(x) &=\left(x^{2}-4 x+2\right) e^{-x} & & f^{\prime \prime}(1) &=-1 / e \\ f^{\prime \prime \prime}(x) &=\left(-x^{2}+6 x-6\right) e^{-x} & & f^{\prime \prime \prime}(1) &=-1 / e \end{aligned} Then \begin{aligned} T_{2}(x) &=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2}(x-a)^{2} \\ &=\frac{1}{e}+\frac{1}{e}(x-1)+\frac{-1 / e}{2}(x-1)^{2} \end{aligned} and \begin{aligned} T_{3}(x) &=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{6}(x-a)^{3} \\ &=\frac{1}{e}+\frac{1}{e}(x-1)+\frac{-1 / e}{2}(x-1)^{2}+\left(\frac{-1 / e}{6}\right)(x-1)^{3} \\ &=\frac{1}{e}+\frac{1}{e}(x-1)-\frac{1}{2 e}(x-1)^{2}-\frac{1}{6 e}(x-1)^{3} \end{aligned}