Question

In Exercises 23–32 we ask you to give Lagrange’s form for the corresponding remainder, $R_4\left(x\right)$

${\tan }^{-1}x$

Short answer

The required answer is $R_4\left(x\right)=\frac{\left\{1-10c^2+5c^4\right\}}{5{(1+c^2)}^5}{x}^5$

Step-by-step solution

Step 1. Given Information  

The given function is$f\left(x\right)={\tan }^{-1}x$

Step 2. Explanation   

Using the Lagrange form for the remainder, we have,

$$\begin{gathered} R_{n+1}\left(x\right)=\frac{f^{n+1}\left(c\right)}{(n+1)!}{x}^{n+1} \\ {R}_4\left(x\right)=\frac{f^5\left(c\right)}{5!}{x}^5 \end{gathered}$$

Now, we will find the fifth derivative of the function,

localid="1649334706302" $$\begin{gathered} f^1\left(x\right)=\frac{1}{1+x^2} \\ {f}^2\left(x\right)=-\frac{2x}{{(1+x^2)}^2} \\ {f}^3\left(x\right)=-\frac{2(1-3x^2)}{{(1+x^2)}^3} \\ {f}^4\left(x\right)=\frac{24(x-x^4)}{{(1+x^2)}^4} \\ {f}^5\left(x\right)=\frac{24}{{(1+x^2)}^5}\left\{1-10x^2+5x^4\right\} \end{gathered}$$

Thus, we get,

localid="1649334781444" role="math" $$\begin{gathered} R_4\left(x\right)=\frac{{\displaystyle \frac{24\left\{1-10c^2+5c^4\right\}}{{(1+c^2)}^5}}}{5!}{x}^5 \\ {R}_4\left(x\right)=\frac{24\left\{1-10c^2+5c^4\right\}}{120{(1+c^2)}^5}{x}^5 \\ {R}_4\left(x\right)=\frac{\left\{1-10c^2+5c^4\right\}}{5{(1+c^2)}^5}{x}^5 \end{gathered}$$