2. the function f has derivatives of all orders for all real numbers and |f^(4)(x)| ≤ 7.5. if the third…

Explanation:

Step1: Recall Lagrange error - bound formula

The Lagrange error - bound formula for the Taylor polynomial of degree $n$ 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 $x$, $n$ is the degree of the Taylor polynomial, $a$ is the center of the Taylor polynomial, and $x$ is the point at which we are approximating the function.

In this case, $n = 3$, $a = 0$, and the interval is $[0,1]$, so $x\in[0,1]$. We know that $|f^{(4)}(c)|\leq7.5$ for all real - valued $c$ (since $|f^{(4)}(x)|\leq7.5$ for all real $x$).

Step2: Substitute values into the formula

Substitute $n = 3$, $a = 0$, and $|f^{(4)}(c)|\leq7.5$ into the Lagrange error - bound formula. We get $|R_3(x)|=\left|\frac{f^{(4)}(c)}{4!}(x - 0)^{4}\right|=\frac{|f^{(4)}(c)|}{24}x^{4}$.

Since $x\in[0,1]$ and $|f^{(4)}(c)|\leq7.5$, to find the maximum value of $|R_3(x)|$ on the interval $[0,1]$, we note that the function $y=\frac{|f^{(4)}(c)|}{24}x^{4}$ is an increasing function of $x$ on the interval $[0,1]$ (because $\frac{|f^{(4)}(c)|}{24}\geq0$).

We substitute $x = 1$ and the maximum value of $|f^{(4)}(c)|$ (which is $7.5$) into the error - bound formula.

$|R_3(1)|=\frac{7.5}{24}\times1^{4}$.

Step3: Calculate the value

$\frac{7.5}{24}=\frac{75}{240}=\frac{5}{16}=0.3125$.

Answer:

$0.3125$