Q. 40 - Calculus

Question

In Exercises  in Section 8.2, you were asked to find the fourth Taylor polynomial ${\mathrm{P}}_4\left(\mathrm{x}\right)$ for the specified function and the given value of ${\mathrm{x}}_0$. In Exercises  give Lagrange’s form for the remainder ${\mathrm{R}}_4\left(\mathrm{x}\right).$

$\sqrt{\mathrm{x}},1$

Step-by-Step Solution

Answer

The Value is ${\mathrm{R}}_4\left(\mathrm{x}\right)=\frac{7}{256}{\mathrm{c}}^{\frac{9}{2}}(\mathrm{x}-1{)}^5$

Step 1: Given information

The function is $\sqrt{\mathrm{x}},1$

Step 2: Simplification

The function's derivatives are $\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}$ is

$$\begin{gathered} \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left[\sqrt{\mathrm{x}}\right] \\ =\frac{1}{2\sqrt{\mathrm{x}}} \end{gathered}$$

That is,

$$\begin{gathered} {\mathrm{f}}^{\text{'}\text{'}}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{1}{2\sqrt{\mathrm{x}}}\right] \\ =\frac{1}{2}\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^{-\frac{1}{2}}\right] \\ =-\frac{1}{2}-\frac{1}{2}{\left(\mathrm{x}\right)}^{-\frac{3}{2}} \\ =-\frac{1}{4}{\mathrm{x}}^{-\frac{3}{2}} \end{gathered}$$

similarly, 

$$\begin{gathered} {\mathrm{f}}^{\text{'}\text{'}}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left[-\frac{1}{4}{\mathrm{x}}^{-\frac{3}{2}}\right] \\ =-\frac{1}{4}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{-\frac{3}{2}} \\ =-\frac{1}{4}-\frac{3}{2}{\mathrm{x}}^{-\frac{5}{2}} \\ =\frac{3}{8}{\mathrm{x}}^{\frac{5}{2}} \end{gathered}$$

Similarly,

$$\begin{gathered} {\mathrm{f}}^{\text{'}\text{'}\text{'}}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{3}{8}{\mathrm{x}}^{\frac{5}{2}}\right] \\ =\frac{3}{8}-\frac{5}{2}{\mathrm{x}}^{\frac{7}{2}} \\ =-\frac{15}{16}{\mathrm{x}}^{\frac{7}{2}} \end{gathered}$$

That is, 

${\mathrm{f}}^{\left(4\right)}\left(\mathrm{x}\right)=-\frac{15}{16}{\mathrm{x}}^{\frac{7}{2}}$

Lastly, 

$$\begin{gathered} {\mathrm{f}}^{\left(3\right)}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left[-\frac{15}{16}{\mathrm{x}}^{\frac{7}{2}}\right] \\ =-\frac{15}{16}\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^{\frac{7}{2}}\right] \\ =-\frac{15}{16}-\frac{7}{2}{\mathrm{x}}^{-\frac{9}{2}} \\ =\frac{105}{32}{\mathrm{x}}^{\frac{9}{2}} \end{gathered}$$

Step 3: Finding Lagrange’s fourth form

If f is a function that may be differentiated $\mathrm{n}+1$ times in some open interval containing the point ${\mathrm{x}}_0$and ${\mathrm{R}}_{\mathrm{n}}\left(\mathrm{x}\right)$ is the nth remainder for $\mathrm{f}$at $\mathrm{x}={\mathrm{x}}_0$, then ${\mathrm{R}}_{\mathrm{n}}\left(\mathrm{x}\right)$ is the ${\mathrm{n}}^{\mathrm{th}}$ remainder for f at $\mathrm{x}={\mathrm{x}}_0$. Then at least one c exists between ${\mathrm{x}}_0$ and x such that ${\mathrm{R}}_{\mathrm{n}}\left(\mathrm{x}\right)=\frac{{\mathrm{f}}^{(\mathrm{n}+1)}\left(\mathrm{c}\right)}{(\mathrm{n}+1)!}{\left(\mathrm{x}-{\mathrm{x}}_0\right)}^{\mathrm{n}+1}$

For , ${\mathrm{f}}^{\left(5\right)}\left(\mathrm{x}\right)=\frac{105}{32}{\mathrm{x}}^{\frac{9}{2}}$ and ${\mathrm{x}}_0=1$ is ${\mathrm{R}}_4\left(\mathrm{x}\right)=\frac{\frac{105}{32}{\mathrm{c}}^{\frac{9}{2}}}{5!}(\mathrm{x}-1{)}^5$

Finally, ${\mathrm{R}}_4\left(\mathrm{x}\right)=\frac{7}{256}{\mathrm{c}}^{\frac{9}{2}}(\mathrm{x}-1{)}^5$