Question

Find the fourth Maclaurin polynomial $P_4\left(x\right)$for the specified function:

$x^2{e}^x$.

Short answer

The fourth Maclaurin polynomial is,

$P_4\left(x\right)=x^2+x^3+\frac{1}{2}{x}^4$.

Step-by-step solution

Step 1. Given Information.

The function is,

$x^2{e}^x$.

Step 2. Describing the polynomial.

Let $f\left(x\right)=x^2{e}^x$.

Since for any function $f$with a derivative of order $4$at $x=0$, the fourth Maclaurin polynomial is given by,

$P_4\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)+\frac{f\text{'}\text{'}\left(0\right)}{2!}{x}^2+\frac{f\text{'}\text{'}\text{'}\left(0\right)}{3!}{x}^3+\frac{f\text{'}\text{'}\text{'}\text{'}\left(0\right)}{4!}{x}^4$.

Step 3. Finding the fourth Maclaurin polynomial.

The value of the function at $x=0$is,

$$\begin{gathered} f\left(0\right)=0^2{e}^0 \\ =0 \end{gathered}$$

The derivatives of the function $f\left(x\right)=x^2{e}^x$are,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d\left(x^2{e}^x\right)}{dx} \\ =x^2\frac{d\left(e^x\right)}{dx}+e^x\frac{d\left(x^2\right)}{dx} \\ =x^2{e}^x+2x{e}^x \\ f\text{'}\left(0\right)=0^2\times e^0+2\times 0\times e^0 \\ =0+0 \\ =0 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{d\left(x^2{e}^x+2x{e}^x\right)}{dx} \\ =\frac{d\left(x^2{e}^x\right)}{dx}+2\frac{d\left(x{e}^x\right)}{dx} \\ =x^2\frac{d\left(e^x\right)}{dx}+e^x\frac{d\left(x^2\right)}{dx}+2\left[x\frac{d\left(e^x\right)}{dx}+e^x\frac{dx}{dx}\right] \\ =x^2{e}^x+2x{e}^x+2x{e}^x+2e^x \\ =x^2{e}^x+4x{e}^x+2e^x \\ f\text{'}\text{'}\left(0\right)=0^2\times e^0+4\times 0\times e^0+2e^0 \\ =0+0+2 \\ =2 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\left(x\right)=\frac{d\left(x^2{e}^x+4x{e}^x+2e^x\right)}{dx} \\ =\frac{d\left(x^2{e}^x\right)}{dx}+4\frac{d\left(x{e}^x\right)}{dx}+2\frac{d\left(e^x\right)}{dx} \\ =x^2\frac{d\left(e^x\right)}{dx}+e^x\frac{d\left(x^2\right)}{dx}+4\left[x\frac{d\left(e^x\right)}{dx}+e^x\frac{dx}{dx}\right]+2e^x \\ =x^2{e}^x+2x{e}^x+4x{e}^x+4e^x+2e^x \\ =x^2{e}^x+6x{e}^x+6e^x \\ f\text{'}\text{'}\text{'}\left(0\right)=0^2\times e^0+6\times 0\times e^0+6\times e^0 \\ =0+0+6 \\ =6 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\text{'}\left(0\right)=\frac{d\left(x^2{e}^x+6x{e}^x+6e^x\right)}{dx} \\ =\frac{d\left(x^2{e}^x\right)}{dx}+6\frac{d\left(x{e}^x\right)}{dx}+6\frac{d\left(e^x\right)}{dx} \\ =x^2\frac{d\left(e^x\right)}{dx}+e^x\frac{d\left(x^2\right)}{dx}+6\left[x\frac{d\left(e^x\right)}{dx}+e^x\frac{dx}{dx}\right]+6e^x \\ =x^2{e}^x+2x{e}^x+6x{e}^x+6e^x+6e^x \\ =x^2{e}^x+8x{e}^x+12e^x \\ f\text{'}\text{'}\text{'}\text{'}\left(0\right)=0^2\times e^0+8\times 0\times e^0+12\times e^0 \\ =12 \end{gathered}$$

The fourth Maclaurin polynomial is,

$$\begin{gathered} P_4\left(x\right)=0+0x+\frac{2}{2!}{x}^2+\frac{6}{3!}{x}^3+\frac{12}{4!}{x}^4 \\ =x^2+x^3+\frac{1}{2}{x}^4 \end{gathered}$$