video example (a) approximate the function f(x) = 5√x by a taylor polynomial of degree 2 at a = 32. (b) how…

video example (a) approximate the function f(x) = 5√x by a taylor polynomial of degree 2 at a = 32. (b) how accurate is this approximation when 31 ≤ x ≤ 33? solution (a) f(x) = 5√x = x^1/5 f(32) = f(x) = f(32) = 1/80 f(x) = -4/25x^(-9/5) f(32) = f(x) = thus the second - degree taylor polynomial is t2(x) = f(32) + f(32)/1! (x - 32) + f(32)/2! (x - 32)^2 = + 1/80(x - 32) - (x - 32)^2. the desired approximation is 5√x ≈ t2(x) =

video example (a) approximate the function f(x) = 5√x by a taylor polynomial of degree 2 at a = 32. (b) how accurate is this approximation when 31 ≤ x ≤ 33? solution (a) f(x) = 5√x = x^1/5 f(32) = f(x) = f(32) = 1/80 f(x) = -4/25x^(-9/5) f(32) = f(x) = thus the second - degree taylor polynomial is t2(x) = f(32) + f(32)/1! (x - 32) + f(32)/2! (x - 32)^2 = + 1/80(x - 32) - (x - 32)^2. the desired approximation is 5√x ≈ t2(x) =

Answer

Explanation:

Step1: Calculate $f(32)$

Given $f(x)=x^{\frac{1}{5}}$, then $f(32)=32^{\frac{1}{5}} = 2$.

Step2: Find the first - derivative $f^{\prime}(x)$

Using the power rule $(x^n)^\prime=nx^{n - 1}$, for $f(x)=x^{\frac{1}{5}}$, we have $f^{\prime}(x)=\frac{1}{5}x^{-\frac{4}{5}}$.

Step3: Calculate $f^{\prime\prime}(32)$

Given $f^{\prime\prime}(x)=-\frac{4}{25}x^{-\frac{9}{5}}$, then $f^{\prime\prime}(32)=-\frac{4}{25}\times32^{-\frac{9}{5}}=-\frac{4}{25}\times\frac{1}{512}=-\frac{1}{3200}$.

Step4: Write out the second - degree Taylor polynomial

The second - degree Taylor polynomial $T_2(x)=f(32)+\frac{f^{\prime}(32)}{1!}(x - 32)+\frac{f^{\prime\prime}(32)}{2!}(x - 32)^2$. Substitute $f(32) = 2$, $f^{\prime}(32)=\frac{1}{80}$, $f^{\prime\prime}(32)=-\frac{1}{3200}$ into the formula: $T_2(x)=2+\frac{1}{80}(x - 32)-\frac{1}{6400}(x - 32)^2$.

Answer:

$f(32)=2$; $f^{\prime}(x)=\frac{1}{5}x^{-\frac{4}{5}}$; $f^{\prime\prime}(32)=-\frac{1}{3200}$; $f^{\prime\prime\prime}(x)=\frac{36}{125}x^{-\frac{14}{5}}$; $T_2(x)=2+\frac{1}{80}(x - 32)-\frac{1}{6400}(x - 32)^2$; $\sqrt[5]{x}\approx2+\frac{1}{80}(x - 32)-\frac{1}{6400}(x - 32)^2$