write a degree 2 taylor polynomial for f(x)= -5 cos(x) centered at x = 4π/3. answer attempt 2 out of 2…

Explanation:

Step1: Recall Taylor - polynomial formula

The degree - 2 Taylor polynomial of a function $f(x)$ centered at $a$ is $P_2(x)=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^2$.

Step2: Find $f(a)$

Given $f(x)=-5\cos(x)$ and $a = \frac{4\pi}{3}$. Then $f(\frac{4\pi}{3})=-5\cos(\frac{4\pi}{3})=-5\times(-\frac{1}{2})=\frac{5}{2}$.

Step3: Find $f^{\prime}(x)$ and $f^{\prime}(a)$

$f^{\prime}(x)=5\sin(x)$. So $f^{\prime}(\frac{4\pi}{3})=5\sin(\frac{4\pi}{3})=5\times(-\frac{\sqrt{3}}{2})=-\frac{5\sqrt{3}}{2}$.

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

$f^{\prime\prime}(x)=5\cos(x)$. Then $f^{\prime\prime}(\frac{4\pi}{3})=5\cos(\frac{4\pi}{3})=5\times(-\frac{1}{2})=-\frac{5}{2}$.

Step5: Substitute into Taylor - polynomial formula

$P_2(x)=f(\frac{4\pi}{3})+f^{\prime}(\frac{4\pi}{3})(x - \frac{4\pi}{3})+\frac{f^{\prime\prime}(\frac{4\pi}{3})}{2}(x - \frac{4\pi}{3})^2$. Substituting the values: $P_2(x)=\frac{5}{2}+\frac{5\sqrt{3}}{2}(x - \frac{4\pi}{3})+\frac{5}{2}(x - \frac{4\pi}{3})^2$.

Answer:

$P_2(x)=\frac{5}{2}+\frac{5\sqrt{3}}{2}(x - \frac{4\pi}{3})+\frac{5}{2}(x - \frac{4\pi}{3})^2$