attempt 1: the answer is correct. find the first - and second - order taylor polynomials of f centered at x…

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of a function $f(x)$ centered at $x = a$ is given by $f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x - a)^n$, where $f^{(n)}(a)$ is the $n$-th derivative of $f$ evaluated at $x=a$. The first - order Taylor polynomial $p_1(x)$ is $f(a)+f^{\prime}(a)(x - a)$ and the second - order Taylor polynomial $p_2(x)$ is $f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^2$.

Step2: Find $f(a)$, $f^{\prime}(x)$ and $f^{\prime}(a)$

Given $f(x)=e^{-4x}$ and $a = 0$. First, $f(0)=e^{-4\times0}=1$. Then, find the first - derivative $f^{\prime}(x)=-4e^{-4x}$, so $f^{\prime}(0)=-4e^{-4\times0}=-4$.

Step3: Calculate the first - order Taylor polynomial

Using the formula $p_1(x)=f(a)+f^{\prime}(a)(x - a)$ with $a = 0$, we have $p_1(x)=f(0)+f^{\prime}(0)x=1+( - 4)x=1 - 4x$.

Step4: Find $f^{\prime\prime}(x)$ and $f^{\prime\prime}(a)$

Differentiate $f^{\prime}(x)=-4e^{-4x}$ to get $f^{\prime\prime}(x)=16e^{-4x}$. Then $f^{\prime\prime}(0)=16e^{-4\times0}=16$.

Step5: Calculate the second - order Taylor polynomial

Using the formula $p_2(x)=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^2$ with $a = 0$, we have $p_2(x)=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2}x^2=1-4x+\frac{16}{2}x^2=1 - 4x+8x^2$.

Answer:

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