3. the function f has derivatives of all orders for all real numbers and f^(5)(x)=e^x. if the fourth…

Explanation:

Step1: Recall Lagrange error - bound formula

The Lagrange error - bound for the $n$th - degree Taylor polynomial $P_n(x)$ of a function $f(x)$ about $x = a$ is given by $R_n(x)=\frac{f^{(n + 1)}(c)}{(n+1)!}(x - a)^{n + 1}$, where $c$ is some number between $a$ and $x$. Here, $n = 4$, $a = 0$, and $x\in[0,1]$, so $R_4(x)=\frac{f^{(5)}(c)}{5!}x^{5}$, and we want to find $\max_{x\in[0,1]}|R_4(x)|$.

Step2: Determine the maximum value of $f^{(5)}(c)$ on $[0,1]$

We are given that $f^{(5)}(x)=e^{x}$. The function $y = e^{x}$ is an increasing function on the real - line. For $c\in[0,1]$, the maximum value of $f^{(5)}(c)$ occurs when $c = 1$, and $f^{(5)}(c)_{\max}=e^{1}=e$.

Step3: Calculate the maximum error

We want to find the maximum of $|R_4(x)|=\frac{f^{(5)}(c)}{5!}x^{5}$ on the interval $[0,1]$. Since $|f^{(5)}(c)|\leq e$ for $c\in[0,1]$ and $x\in[0,1]$, we substitute $f^{(5)}(c)=e$ and $x = 1$ into the error formula. Then $|R_4(x)|{\max}=\frac{e}{5!}$. Since $5!=5\times4\times3\times2\times1 = 120$, we have $|R_4(x)|{\max}=\frac{e}{120}$.

Answer:

$\frac{e}{120}$