3. the function f has derivatives of all orders for all real numbers and f^(4)(x) = 1/8 e^(x/2). if the…

Explanation:

Step1: Recall Lagrange error formula

The Lagrange error formula for the $n$-th - degree Taylor polynomial $P_n(x)$ of a function $f(x)$ is $R_n(x)=\frac{f^{(n + 1)}(c)}{(n+1)!}(x - a)^ {n+1}$, where $c$ is some number between $a$ and $x$. Here, $n = 3$, $a = 0$, and $x\in[0,1]$, so $R_3(x)=\frac{f^{(4)}(c)}{4!}x^{4}$, and we want to find $\max_{x\in[0,1]}|R_3(x)|$.

Step2: Find an upper - bound for $|f^{(4)}(c)|$

We are given that $f^{(4)}(x)=\frac{1}{8}e^{\frac{x}{2}}$. The function $y = f^{(4)}(x)$ is an increasing function on the interval $[0,1]$ since its derivative $y'=\frac{1}{16}e^{\frac{x}{2}}>0$ for all real $x$. So, on the interval $[0,1]$, the maximum value of $|f^{(4)}(c)|$ occurs when $c = 1$. Then $|f^{(4)}(c)|_{\max}=|f^{(4)}(1)|=\frac{1}{8}e^{\frac{1}{2}}$.

Step3: Calculate the error bound

We know that $|R_3(x)|\leq\frac{|f^{(4)}(c)|{\max}}{4!}x^{4}$. Since $x\in[0,1]$, the maximum value of $|R_3(x)|$ occurs when $x = 1$. Substituting $|f^{(4)}(c)|{\max}=\frac{1}{8}e^{\frac{1}{2}}$ and $x = 1$ into the error - bound formula, we have $|R_3|_{\max}=\frac{\frac{1}{8}e^{\frac{1}{2}}}{24}=\frac{e^{\frac{1}{2}}}{192}$.

Answer:

$\frac{\sqrt{e}}{192}$