use the degree 2 maclaurin polynomial of f(x) = cos(2x) to approximate cos(2). (enter a number for your…

use the degree 2 maclaurin polynomial of f(x) = cos(2x) to approximate cos(2). (enter a number for your answer.) question 9 (1 point) a 5 - th degree taylor polynomial of a function f(x) at x = 0 could potentially have the form (select all that apply): t_5(x)=x^5 t_5(x)=x t_5(x)=x^6 t_5(x)=1 + x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}

use the degree 2 maclaurin polynomial of f(x) = cos(2x) to approximate cos(2). (enter a number for your answer.) question 9 (1 point) a 5 - th degree taylor polynomial of a function f(x) at x = 0 could potentially have the form (select all that apply): t_5(x)=x^5 t_5(x)=x t_5(x)=x^6 t_5(x)=1 + x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}

Answer

Explanation:

Step1: Recall Mac - Laurin formula

The Mac - Laurin series of a function $f(x)$ is given by $f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\frac{f^{(3)}(0)}{3!}x^{3}+\cdots$. For $y = f(x)=\cos(2x)$, we first find its derivatives and evaluate at $x = 0$.

Step2: Find derivatives and their values at $x = 0$

  • $f(x)=\cos(2x)$, so $f(0)=\cos(0)=1$.
  • $f^{\prime}(x)=-2\sin(2x)$, then $f^{\prime}(0)=-2\sin(0)=0$.
  • $f^{\prime\prime}(x)=-4\cos(2x)$, and $f^{\prime\prime}(0)=-4\cos(0)=-4$.

Step3: Write degree - 2 Mac - Laurin polynomial

The degree - 2 Mac - Laurin polynomial $P_2(x)$ of $y = f(x)$ is $P_2(x)=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^{2}$. Substituting the values we found: $P_2(x)=1 + 0\times x+\frac{-4}{2}x^{2}=1 - 2x^{2}$.

Step4: Approximate $\cos(2)$

To approximate $\cos(2)$, we substitute $x = 1$ into $P_2(x)$ (since when approximating $\cos(2)=\cos(2\times1)$), so $P_2(1)=1-2\times1^{2}=1 - 2=-1$.

For Question 9:

A $5$-th degree Taylor polynomial of a function $f(x)$ at $x = 0$ has the form $T_5(x)=\sum_{n = 0}^{5}\frac{f^{(n)}(0)}{n!}x^{n}=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\frac{f^{(3)}(0)}{3!}x^{3}+\frac{f^{(4)}(0)}{4!}x^{4}+\frac{f^{(5)}(0)}{5!}x^{5}$.

  • Option 1: $T_5(x)=x^{5}$ is possible when $f(0)=f^{\prime}(0)=f^{\prime\prime}(0)=f^{(3)}(0)=f^{(4)}(0)=0$ and $f^{(5)}(0)=5!$.
  • Option 2: $T_5(x)=x$ is a first - degree polynomial, not a fifth - degree polynomial.
  • Option 3: $T_5(x)=x^{6}$ is a sixth - degree polynomial, not a fifth - degree polynomial.
  • Option 4: $T_5(x)=1 + x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\frac{x^{5}}{5!}$ is in the general form of a fifth - degree Taylor polynomial at $x = 0$.

Answer:

For the first part: -1 For Question 9: A. $T_5(x)=x^{5}$ D. $T_5(x)=1 + x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\frac{x^{5}}{5!}$