Approximate f(x)=(2x²+3x)e^-x by its degree-4 Taylor

Approximate $f(x)=(2x^2+3x)e^{-x}$ by its degree-4 Taylor (Maclaurin) polynomial about $x=0$ and estimate the error for $-\tfrac1{10}\lt x\lt \tfrac1{10}$.

Step 1: Multiply $(2x^2+3x)$ into the series for $e^{-x}$

With $e^{-x}=1-x+\tfrac{x^2}2-\tfrac{x^3}6+\tfrac{x^4}{24}+\cdots$, collecting by degree through $x^4$: \[x^1:\,3x;\quad x^2:\,-3x^2+2x^2=-x^2;\quad x^3:\,\tfrac32x^3-2x^3=-\tfrac12x^3;\quad x^4:\,-\tfrac12x^4+x^4=\tfrac12x^4. \tag{1}\]

Step 2: Degree-4 polynomial

\[P_4(x) = 3x-x^2-\frac{x^3}2+\frac{x^4}2\]

Step 3: Fifth derivative via the Leibniz rule

Writing $f=(2x^2+3x)e^{-x}$, with $(2x^2+3x)^{(0)}=2x^2+3x$, $(2x^2+3x)'=4x+3$, $(2x^2+3x)''=4$, and $(e^{-x})^{(k)}=(-1)^ke^{-x}$: \[f^{(5)}(x) = \bigl[-(2x^2+3x)+5(4x+3)-40\bigr]e^{-x} = (-2x^2+17x-25)\,e^{-x}. \tag{2}\]

Step 4: Lagrange remainder and its bound for $|x|\lt \tfrac1{10}$

\[ R_4(x) = \frac{f^{(5)}(\xi)}{5!}x^5 = \frac{(-2\xi^2+17\xi-25)e^{-\xi}}{120}x^5. \tag{3} \]

For $|\xi|\lt \tfrac1{10}$: $|-2\xi^2+17\xi-25|\lt 26.72$ and $e^{-\xi}\lt e^{1/10}\lt 1.11$, so $|f^{(5)}(\xi)|\lt 30$, giving

\[ |R_4(x)| \lt \frac{30}{120}\left(\frac1{10}\right)^5 = 2.5\times10^{-6}. \tag{4} \]

Step 5: Verification

\[x=0.1\text{: }P_4(0.1)=0.28955,\ f(0.1)=(0.02+0.3)e^{-0.1}\approx0.289548,\ |{\rm error}|\approx2.0\times10^{-6}\lt 2.5\times10^{-6}\ \checkmark\]