Question

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

$\ln (1+x)$

Short answer

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

Step-by-step solution

Step 1. Given Information   

The given function is$f\left(x\right)=\ln (1+x)$

Step 2. Explanation  

Using the Lagrange form for the remainder, we have,

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

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

$$\begin{gathered} f^1\left(x\right)=\frac{1}{1+x} \\ {f}^2\left(x\right)=-\frac{1}{{(1+x)}^2} \\ {f}^3\left(x\right)=\frac{2}{{(1+x)}^3} \\ {f}^4\left(x\right)=-\frac{6}{{(1+x)}^4} \\ {f}^5\left(x\right)=\frac{24}{{(1+x)}^5} \end{gathered}$$

Thus, we get,

$$\begin{gathered} R_4\left(x\right)=\frac{{\displaystyle \frac{24}{{(1+c)}^5}}}{5!}{x}^5 \\ {R}_4\left(x\right)=\frac{24}{120{(1+c)}^5}{x}^5 \\ {R}_4\left(x\right)=\frac{1}{5{(1+c)}^5}{x}^5 \end{gathered}$$