attempt 1: 10 attempts remaining. suppose g is a function which has continuous derivatives, and that g(3)=…

attempt 1: 10 attempts remaining. suppose g is a function which has continuous derivatives, and that g(3)= -1, g(3)= -4, g(3)= -5, g(3)= -5. (a) what is the taylor polynomial of degree 2 for g centered at a = 3? p2(x)= (b) what is the taylor polynomial of degree 3 for g centered at a = 3? p3(x)= (c) use the two polynomials that you found in parts (a) and (b) to approximate g(3.1). (do not round until the final answer. then round to six decimal places as needed.) with p2, g(3.1)≈ with p3, g(3.1)≈ submit answer next item

attempt 1: 10 attempts remaining. suppose g is a function which has continuous derivatives, and that g(3)= -1, g(3)= -4, g(3)= -5, g(3)= -5. (a) what is the taylor polynomial of degree 2 for g centered at a = 3? p2(x)= (b) what is the taylor polynomial of degree 3 for g centered at a = 3? p3(x)= (c) use the two polynomials that you found in parts (a) and (b) to approximate g(3.1). (do not round until the final answer. then round to six decimal places as needed.) with p2, g(3.1)≈ with p3, g(3.1)≈ submit answer next item

Answer

Explanation:

Step1: Recall Taylor - polynomial formula

The Taylor polynomial of degree $n$ for a function $g(x)$ centered at $a$ is given by $p_n(x)=\sum_{k = 0}^{n}\frac{g^{(k)}(a)}{k!}(x - a)^k=g(a)+g^{\prime}(a)(x - a)+\frac{g^{\prime\prime}(a)}{2!}(x - a)^2+\cdots+\frac{g^{(n)}(a)}{n!}(x - a)^n$.

Step2: Find $p_2(x)$

For $n = 2$, $a = 3$, $g(3)=-1$, $g^{\prime}(3)=-4$, $g^{\prime\prime}(3)=-5$. [ \begin{align*} p_2(x)&=g(3)+g^{\prime}(3)(x - 3)+\frac{g^{\prime\prime}(3)}{2}(x - 3)^2\ &=-1-4(x - 3)-\frac{5}{2}(x - 3)^2 \end{align*} ]

Step3: Find $p_3(x)$

For $n = 3$, $a = 3$, we use the Taylor - polynomial formula. [ \begin{align*} p_3(x)&=g(3)+g^{\prime}(3)(x - 3)+\frac{g^{\prime\prime}(3)}{2}(x - 3)^2+\frac{g^{\prime\prime\prime}(3)}{6}(x - 3)^3\ &=-1-4(x - 3)-\frac{5}{2}(x - 3)^2-\frac{5}{6}(x - 3)^3 \end{align*} ]

Step4: Approximate $g(3.1)$ using $p_2(x)$

Let $x = 3.1$, then $x-3=0.1$. [ \begin{align*} p_2(3.1)&=-1-4\times(3.1 - 3)-\frac{5}{2}\times(3.1 - 3)^2\ &=-1-4\times0.1-\frac{5}{2}\times0.01\ &=-1 - 0.4-0.025\ &=-1.425000 \end{align*} ]

Step5: Approximate $g(3.1)$ using $p_3(x)$

Let $x = 3.1$, then $x - 3=0.1$. [ \begin{align*} p_3(3.1)&=-1-4\times(3.1 - 3)-\frac{5}{2}\times(3.1 - 3)^2-\frac{5}{6}\times(3.1 - 3)^3\ &=-1-4\times0.1-\frac{5}{2}\times0.01-\frac{5}{6}\times0.001\ &=-1-0.4 - 0.025-\frac{5}{6000}\ &=-1-0.4-0.025 - 0.000833\ &=-1.425833 \end{align*} ]

Answer:

(a) $p_2(x)=-1-4(x - 3)-\frac{5}{2}(x - 3)^2$ (b) $p_3(x)=-1-4(x - 3)-\frac{5}{2}(x - 3)^2-\frac{5}{6}(x - 3)^3$ (c) With $p_2$, $g(3.1)\approx - 1.425000$ With $p_3$, $g(3.1)\approx - 1.425833$