91Ó°ÊÓ

${\left(x^2+1\right)}^{3/2},x_0=0$

The formula used:$P_3\left(x\right)=f\left(0\right)+f^{\text{'}}\left(0\right)x+\frac{f^{\text{'}\text{'}}\left(0\right)}{2!}{x}^2+\frac{f^{\text{'}\text{'}\text{'}}\left(0\right)}{3!}{x}^3$

Consider the function $f\left(x\right)={\left(x^2+1\right)}^{\frac{3}{2}}$

Since any function $f$with a derivative of order $3$at $x=0$the third Maclaurin polynomial is given by

$P_3\left(x\right)=f\left(0\right)+f^{\text{'}}\left(0\right)x+\frac{f^{\text{'}\text{'}}\left(0\right)}{2!}{x}^2+\frac{f^{\text{'}\text{'}\text{'}}\left(0\right)}{3!}{x}^3$

Therefore, first, find the value of the function along with $f^{\text{'}}\left(x\right),f^{\text{'}\text{'}}\left(x\right)$and $f^{\text{'}\text{'}\text{'}}\left(x\right)$at $x=0$

Thus, the value of the function $x=0$is

The derivatives of the function $f\left(x\right)={\left(x^2+1\right)}^{\frac{3}{2}}$are

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}{\left(x^2+1\right)}^{\frac{3}{2}} \\ =\frac{3}{2}{\left(x^2+1\right)}^{\frac{1}{2}}·2x \\ =3x{\left(x^2+1\right)}^{\frac{1}{2}} \end{gathered}$$

So, at $x=0$

$$\begin{gathered} f^{\text{'}}\left(0\right)=3·0·{\left(0^2+1\right)}^{\frac{1}{2}} \\ =0 \end{gathered}$$

Also,

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\left[3x{\left(x^2+1\right)}^{\frac{1}{2}}\right] \\ =3\left\{x\frac{d}{dx}{\left(x^2+1\right)}^{\frac{1}{2}}+{\left(x^2+1\right)}^{\frac{1}{2}}\frac{d}{dx}\left[x\right]\right\} \\ =3\left[x·\frac{1}{2}{\left(x^2+1\right)}^{-\frac{1}{2}}·2x+{\left(x^2+1\right)}^{\frac{1}{2}}+1\right] \\ =3\left[x^2{\left(x^2+1\right)}^{-\frac{1}{2}}+{\left(x^2+1\right)}^{\frac{1}{2}}\right] \end{gathered}$$

So, at $x=0$

So, at $x=0$

$$\begin{gathered} f^{\text{'}\text{'}}\left(0\right)=3\left[0^2{\left(0^2+1\right)}^{-\frac{1}{2}}+{\left(0^2+1\right)}^{\frac{1}{2}}\right] \\ =3(0+1) \\ =3 \end{gathered}$$

Again

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=3\frac{d}{dx}\left[x^2{\left(x^2+1\right)}^{\frac{1}{2}}+{\left(x^2+1\right)}^{\frac{1}{2}}\right] \\ =3\left\{x^2\frac{d}{dx}{\left(x^2+1\right)}^{-\frac{1}{2}}+{\left(x^2+1\right)}^{-\frac{1}{2}}\frac{d}{dx}\left[x^2\right]+\frac{d}{dx}{\left(x^2+1\right)}^{\frac{1}{2}}\right\} \\ =3\left\{x^2·-\frac{1}{2}{\left(x^2+1\right)}^{\frac{-3}{2}}·2x+{\left(x^2+1\right)}^{-\frac{1}{2}}·2x+\frac{1}{2}{\left(x^2+1\right)}^{\frac{1}{2}}·2x\right\} \\ =3\left[-x^3{\left(x^2+1\right)}^{-\frac{3}{2}}+2x{\left(x^2+1\right)}^{-\frac{1}{2}}+x{\left(x^2+1\right)}^{-\frac{1}{2}}\right] \end{gathered}$$

So, at $x=0$

$$\begin{gathered} f^{\text{'}\text{'}}\left(0\right)=3\left[-0^3{\left(0^2+1\right)}^{-\frac{3}{2}}+2·0·{\left(0^2+1\right)}^{-\frac{1}{2}}+0{\left(0^2+1\right)}^{\frac{1}{2}}\right] \\ =0 \end{gathered}$$

As a result, the third-order Maclaurin series for the function $f\left(x\right)={\left(x^2+1\right)}^{\frac{3}{2}}$is

$P_3\left(x\right)=1+0·x+\frac{3}{2!}{x}^2+\frac{0}{3!}{x}^3$

Implies that

$P_3\left(x\right)=1+\frac{3}{2}{x}^2$