question write a degree 3 taylor polynomial for f(x)=-3 sin(-5x) centered at x = 11π/6. answer attempt 1 out…

Explanation:

Step1: Recall Taylor - series formula

The Taylor polynomial of degree $n$ for a function $f(x)$ centered at $a$ is $P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^2+\frac{f^{(3)}(a)}{3!}(x - a)^3$. First, find $f(x)=- 3\sin(-5x)=3\sin(5x)$, and $a=\frac{11\pi}{6}$.

Step2: Calculate $f(a)$

$f(x)=3\sin(5x)$, so $f(\frac{11\pi}{6})=3\sin(\frac{55\pi}{6})=3\sin(9\pi+\frac{\pi}{6})=-3\sin(\frac{\pi}{6})=-\frac{3}{2}$.

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

$f^{\prime}(x)=15\cos(5x)$, then $f^{\prime}(\frac{11\pi}{6})=15\cos(\frac{55\pi}{6})=15\cos(9\pi+\frac{\pi}{6})=-15\cos(\frac{\pi}{6})=-\frac{15\sqrt{3}}{2}$.

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

$f^{\prime\prime}(x)=-75\sin(5x)$, so $f^{\prime\prime}(\frac{11\pi}{6})=-75\sin(\frac{55\pi}{6})=-75\sin(9\pi+\frac{\pi}{6}) = 75\sin(\frac{\pi}{6})=\frac{75}{2}$.

Step5: Calculate $f^{(3)}(x)$ and $f^{(3)}(a)$

$f^{(3)}(x)=-375\cos(5x)$, then $f^{(3)}(\frac{11\pi}{6})=-375\cos(\frac{55\pi}{6})=-375\cos(9\pi+\frac{\pi}{6})=375\cos(\frac{\pi}{6})=\frac{375\sqrt{3}}{2}$.

Step6: Construct $P_3(x)$

$P_3(x)=f(\frac{11\pi}{6})+f^{\prime}(\frac{11\pi}{6})(x - \frac{11\pi}{6})+\frac{f^{\prime\prime}(\frac{11\pi}{6})}{2}(x - \frac{11\pi}{6})^2+\frac{f^{(3)}(\frac{11\pi}{6})}{6}(x - \frac{11\pi}{6})^3$ $=-\frac{3}{2}-\frac{15\sqrt{3}}{2}(x - \frac{11\pi}{6})+\frac{75}{4}(x - \frac{11\pi}{6})^2+\frac{125\sqrt{3}}{4}(x - \frac{11\pi}{6})^3$.

Answer:

$-\frac{3}{2}-\frac{15\sqrt{3}}{2}(x - \frac{11\pi}{6})+\frac{75}{4}(x - \frac{11\pi}{6})^2+\frac{125\sqrt{3}}{4}(x - \frac{11\pi}{6})^3$