2. the function f has derivatives of all orders for all real numbers and |f^(4)(x)| ≤ 1.8. if the third…

Explanation:

Step1: Recall Lagrange error - bound formula

The Lagrange error - bound formula for the $n$ - th degree Taylor polynomial $P_n(x)$ of a function $f(x)$ is given by $|R_n(x)|=\left|\frac{f^{(n + 1)}(c)}{(n+1)!}(x - a)^{n + 1}\right|$, where $c$ is some number between $a$ and $x$, $a$ is the center of the Taylor polynomial, and $n$ is the degree of the Taylor polynomial.

In this case, $n = 3$, $a = 0$, and $x\in[0,1]$. We know that $|f^{(4)}(x)|\leq1.8$ for all real $x$.

Step2: Substitute values into the formula

We want to find the maximum of $|R_3(x)|$ on the interval $[0,1]$. Since $a = 0$, the formula becomes $|R_3(x)|=\left|\frac{f^{(4)}(c)}{4!}x^{4}\right|$, where $c$ is between $0$ and $x$.

We know that $|f^{(4)}(c)|\leq1.8$ and $x\in[0,1]$. To find the maximum value of $|R_3(x)|$ on $[0,1]$, we note that the function $y = x^{4}$ on $[0,1]$ has a maximum value of $1$ when $x = 1$.

So, $|R_3(x)|\leq\frac{\max|f^{(4)}(c)|}{4!}\times\max|x^{4}|$.

Step3: Calculate the error - bound

We know that $\max|f^{(4)}(c)| = 1.8$ and $\max|x^{4}|=1$ (for $x\in[0,1]$), and $4!=24$.

Then $|R_3(x)|\leq\frac{1.8}{24}=0.075$.

Answer:

$0.075$