Q. 38 - Calculus

Question

In Exercises 41–48 in Section 8.2, you were asked to find the fourth Taylor polynomial $P_4\left(x\right)$ for the specified function and the given value of x 0. Here give Lagrange’s form for the remainder $R_4\left(x\right)$.  

      $e^x, 1$

Step-by-Step Solution

Answer

Ans:   $R_4\left(x\right)=\frac{e^c}{120}(x-1{)}^5$

Step 1. Given information.

given,

        $e^x, 1$

Step 2. Consider the given function,

 The Lagrange's form for the remainder is $R_n\left(x\right)=\frac{f^{(n+1)}\left(c\right)}{(n+1)!}{\left(x-x_0\right)}^{n+1}$, where c is between $x_{0 }and x$.

Since $f\left(x\right)=e^x$, so we have $f^{(n+1)}\left(c\right)=e^c\text{ for every }n\ge 0$ 

Also, the series of Taylor, so use $x_0=1$

Therefore, 

     

Since $f^{\left(5\right)}\left(x\right)=e^x\text{ and }{x}_0=1\text{ then }$

     $R_4\left(x\right)=\frac{f^5\left(c\right)}{5!}(x-1{)}^5$

That is, 

      $R_4\left(x\right)=\frac{e^c}{120}(x-1{)}^5$