question number 21. use the values in the table below and the formula for taylor polynomials to give the 4th…

Explanation:

Step1: Recall Taylor - polynomial formula

The $n$ -th degree Taylor polynomial of a function $f(x)$ centered at $a$ is given by $P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k$. When $a = 0$, it becomes $P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^k$. For $n = 4$, we have $P_4(x)=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+\frac{f^{(4)}(0)}{4!}x^4$.

Step2: Substitute the given values

We are given $f(0)=-2$, $f^{\prime}(0)=2$, $f^{\prime\prime}(0)=3$, $f^{(3)}(0)=4$, $f^{(4)}(0)=1$. Substituting these values into the formula: [ \begin{align*} P_4(x)&=-2 + 2x+\frac{3}{2!}x^2+\frac{4}{3!}x^3+\frac{1}{4!}x^4\ &=-2 + 2x+\frac{3}{2}x^2+\frac{4}{6}x^3+\frac{1}{24}x^4\ &=-2 + 2x+\frac{3}{2}x^2+\frac{2}{3}x^3+\frac{1}{24}x^4 \end{align*} ]

Answer:

$P_4(x)=-2 + 2x+\frac{3}{2}x^2+\frac{2}{3}x^3+\frac{1}{24}x^4$