a graphing calculator is recommended. find the taylor polynomial $t_3(x)$ for the function $f$ centered at…

Explanation:

Step1: Recall Taylor - polynomial formula

The Taylor polynomial of degree $n$ for a function $f(x)$ centered at $a$ is given by $T_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^2+\frac{f^{(3)}(a)}{3!}(x - a)^3+\cdots+\frac{f^{(n)}(a)}{n!}(x - a)^n$. Here $n = 3$ and $a=\frac{\pi}{2}$.

Step2: Find the function and its derivatives at $x = a$

• First, find $f(x)=\cos(x)$. Then $f\left(\frac{\pi}{2}\right)=\cos\left(\frac{\pi}{2}\right)=0$.
• Next, find the first - derivative $f^{\prime}(x)=-\sin(x)$. So $f^{\prime}\left(\frac{\pi}{2}\right)=-\sin\left(\frac{\pi}{2}\right)=- 1$.
• Then, find the second - derivative $f^{\prime\prime}(x)=-\cos(x)$. So $f^{\prime\prime}\left(\frac{\pi}{2}\right)=-\cos\left(\frac{\pi}{2}\right)=0$.
• Finally, find the third - derivative $f^{(3)}(x)=\sin(x)$. So $f^{(3)}\left(\frac{\pi}{2}\right)=\sin\left(\frac{\pi}{2}\right)=1$.

Step3: Construct the Taylor polynomial $T_3(x)$

Substitute the values into the Taylor - polynomial formula: [ \begin{align*} T_3(x)&=f\left(\frac{\pi}{2}\right)+f^{\prime}\left(\frac{\pi}{2}\right)\left(x-\frac{\pi}{2}\right)+\frac{f^{\prime\prime}\left(\frac{\pi}{2}\right)}{2!}\left(x - \frac{\pi}{2}\right)^2+\frac{f^{(3)}\left(\frac{\pi}{2}\right)}{3!}\left(x-\frac{\pi}{2}\right)^3\ &=0+(-1)\left(x-\frac{\pi}{2}\right)+\frac{0}{2}\left(x - \frac{\pi}{2}\right)^2+\frac{1}{6}\left(x-\frac{\pi}{2}\right)^3\ &=-(x - \frac{\pi}{2})+\frac{1}{6}(x - \frac{\pi}{2})^3 \end{align*} ]

Answer:

$-(x-\frac{\pi}{2})+\frac{1}{6}(x - \frac{\pi}{2})^3$