attempt 1: 10 attempts remaining. find the first - and second - order taylor polynomials of f centered at x…

attempt 1: 10 attempts remaining. find the first - and second - order taylor polynomials of f centered at x = a. use a graphing utility to graph f and the two approximations on the same screen. f(x)=e^(-4x), a = 0 enter the taylor polynomials below. p1(x)= p2(x)= submit answer next item

attempt 1: 10 attempts remaining. find the first - and second - order taylor polynomials of f centered at x = a. use a graphing utility to graph f and the two approximations on the same screen. f(x)=e^(-4x), a = 0 enter the taylor polynomials below. p1(x)= p2(x)= submit answer next item

Answer

Explanation:

Step1: Recall Taylor - polynomial formula

The $n$-th Taylor polynomial of a function $f(x)$ centered at $x = a$ is given by $P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k$, where $f^{(k)}(x)$ is the $k$-th derivative of $f(x)$.

Step2: Find the function value and its first - derivative at $a = 0$

First, find $f(x)=e^{-4x}$. Then $f(0)=e^{-4\times0}=1$. The first - derivative of $f(x)$ using the chain - rule: $f^\prime(x)=-4e^{-4x}$, and $f^\prime(0)=-4e^{-4\times0}=-4$. The second - derivative of $f(x)$: $f^{\prime\prime}(x)=(-4)\times(-4)e^{-4x}=16e^{-4x}$, and $f^{\prime\prime}(0)=16e^{-4\times0}=16$.

Step3: Calculate the first - order Taylor polynomial $p_1(x)$

For $n = 1$, $P_1(x)=\frac{f(0)}{0!}(x - 0)^0+\frac{f^\prime(0)}{1!}(x - 0)^1$. Since $f(0) = 1$ and $f^\prime(0)=-4$, then $p_1(x)=1-4x$.

Step4: Calculate the second - order Taylor polynomial $p_2(x)$

For $n = 2$, $P_2(x)=\frac{f(0)}{0!}(x - 0)^0+\frac{f^\prime(0)}{1!}(x - 0)^1+\frac{f^{\prime\prime}(0)}{2!}(x - 0)^2$. Substitute $f(0) = 1$, $f^\prime(0)=-4$, and $f^{\prime\prime}(0)=16$ into the formula: $p_2(x)=1-4x+\frac{16}{2}x^2=1 - 4x+8x^2$.

Answer:

$p_1(x)=1 - 4x$ $p_2(x)=1 - 4x+8x^2$