attempt 1: 10 attempts remaining. (a) find the second - order taylor polynomial of f(x)=√(x/3) centered at a…

attempt 1: 10 attempts remaining. (a) find the second - order taylor polynomial of f(x)=√(x/3) centered at a = 3. p2(x)= (b) use the result obtained in part (a) to approximate √0.98 and compare your approximation to that produced directly by your calculating utility. (do not round until the final answer. then round to four decimal places as needed.) √0.98≈ submit answer next item

attempt 1: 10 attempts remaining. (a) find the second - order taylor polynomial of f(x)=√(x/3) centered at a = 3. p2(x)= (b) use the result obtained in part (a) to approximate √0.98 and compare your approximation to that produced directly by your calculating utility. (do not round until the final answer. then round to four decimal places as needed.) √0.98≈ submit answer next item

Answer

Explanation:

Step1: Recall Taylor - polynomial formula

The second - order Taylor polynomial of a function $f(x)$ centered at $a$ is given by $p_2(x)=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^2$.

Step2: First, find $f(a)$

Given $f(x)=\sqrt{\frac{x}{3}}=\frac{1}{\sqrt{3}}x^{\frac{1}{2}}$, and $a = 3$. Then $f(3)=\sqrt{\frac{3}{3}} = 1$.

Step3: Find the first - derivative $f^{\prime}(x)$

Using the power rule $(x^n)^\prime=nx^{n - 1}$, $f^{\prime}(x)=\frac{1}{\sqrt{3}}\times\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{3x}}$. So $f^{\prime}(3)=\frac{1}{2\sqrt{3\times3}}=\frac{1}{6}$.

Step4: Find the second - derivative $f^{\prime\prime}(x)$

$f^{\prime\prime}(x)=\frac{1}{2\sqrt{3}}\times(-\frac{1}{2})x^{-\frac{3}{2}}=-\frac{1}{4\sqrt{3x^{3}}}$. Then $f^{\prime\prime}(3)=-\frac{1}{4\sqrt{3\times3^{3}}}=-\frac{1}{36}$.

Step5: Construct the second - order Taylor polynomial

$p_2(x)=f(3)+f^{\prime}(3)(x - 3)+\frac{f^{\prime\prime}(3)}{2}(x - 3)^2$. Substitute $f(3) = 1$, $f^{\prime}(3)=\frac{1}{6}$, and $f^{\prime\prime}(3)=-\frac{1}{36}$ into the formula: $p_2(x)=1+\frac{1}{6}(x - 3)-\frac{1}{72}(x - 3)^2$.

Step6: Approximate $\sqrt{0.98}$

We want to approximate $\sqrt{0.98}$. Let $\frac{x}{3}=0.98$, then $x = 2.94$. Substitute $x = 2.94$ into $p_2(x)$: [ \begin{align*} p_2(2.94)&=1+\frac{1}{6}(2.94 - 3)-\frac{1}{72}(2.94 - 3)^2\ &=1+\frac{1}{6}\times(-0.06)-\frac{1}{72}\times(-0.06)^2\ &=1-0.01-\frac{1}{72}\times0.0036\ &=1 - 0.01-0.00005\ &=0.98995\approx0.9900 \end{align*} ] Using a calculator, $\sqrt{0.98}\approx0.9899$.

Answer:

(a) $p_2(x)=1+\frac{1}{6}(x - 3)-\frac{1}{72}(x - 3)^2$ (b) $\sqrt{0.98}\approx0.9900$