question 8 of 10 > find the taylor polynomial t₂(x) and compute the error |f(x) - t₂(x)| for the given…

Explanation:

Step1: Recall Taylor - polynomial formula

The second - degree Taylor polynomial $T_2(x)$ of a function $f(x)$ centered at $a$ is given by $T_2(x)=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^2$. First, find the derivatives of $f(x)=\cos(x)$. The first - derivative $f^{\prime}(x)=-\sin(x)$ and the second - derivative $f^{\prime\prime}(x)=-\cos(x)$.

Step2: Evaluate $f(a), f^{\prime}(a), f^{\prime\prime}(a)$ at $a = 0$

When $a = 0$, $f(0)=\cos(0)=1$, $f^{\prime}(0)=-\sin(0)=0$, $f^{\prime\prime}(0)=-\cos(0)=-1$.

Step3: Write out the Taylor polynomial $T_2(x)$

Substitute into the Taylor - polynomial formula: $T_2(x)=1+0\times(x - 0)+\frac{-1}{2}(x - 0)^2=1-\frac{1}{2}x^2$.

Step4: Evaluate $T_2(\frac{\pi}{6})$

Substitute $x = \frac{\pi}{6}$ into $T_2(x)$: $T_2(\frac{\pi}{6})=1-\frac{1}{2}(\frac{\pi}{6})^2=1-\frac{\pi^2}{72}\approx1 - \frac{9.869604}{72}\approx1 - 0.137078=0.862922$.

Step5: Evaluate $f(\frac{\pi}{6})$

$f(\frac{\pi}{6})=\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\approx0.866025$.

Step6: Compute the error $|f(\frac{\pi}{6})-T_2(\frac{\pi}{6})|$

$|f(\frac{\pi}{6})-T_2(\frac{\pi}{6})|=\left|\frac{\sqrt{3}}{2}-(1 - \frac{\pi^2}{72})\right|\approx|0.866025 - 0.862922|=0.003103$.

Answer:

$T_2(\frac{\pi}{6})\approx0.862922$ $|f(\frac{\pi}{6})-T_2(\frac{\pi}{6})|\approx0.003103$