Question

In Exercises 41–48 find the fourth Taylor polynomial $P_4\left(x\right)$for the specified function and the given value of $x_0$

$46.\sqrt[3]{x},1$

Short answer

The fourth Taylor polynomial of the function $f\left(x\right)=\sqrt[3]{x}$at $x=1$is $P_4\left(x\right)=1+\frac{1}{3}(x-1)-\frac{1}{9}(x-1{)}^2+\frac{5}{81}(x-1{)}^3-\frac{10}{243}(x-1{)}^4$

Step-by-step solution

Step 1. Given data

We have the given function $f\left(x\right)=\sqrt[3]{x}$with a derivative of order $4$at$x=1$.

Step 2. The fourth taylor polynomial

The fourth taylor polynomial for $x=1$is given by,

$P_4\left(x\right)=f\left(1\right)+f^{\mathrm{\prime }}\left(1\right)(x-1)+\frac{f^{\prime \prime }\left(1\right)}{2!}(x-1{)}^2+\frac{f^{\prime \prime }\left(1\right)}{3!}(x-1{)}^3+\frac{f^{\prime \prime \prime \prime }\left(1\right)}{4!}(x-1{)}^4$

Therefore, we have to find the value of the function along with $f^{\text{'}}\left(x\right),f^{\text{'}\text{'}}\left(x\right),f^{\text{'}\text{'}\text{'}}\left(x\right)$and$f\text{'}\text{'}\text{'}\text{'}\left(x\right)$at $x=1$.

The value of the function atlocalid="1649396841451" $x=1$islocalid="1649396836147" $f\left(1\right)=\sqrt[3]{1}=1$

Step 3. Find f'(x)

The derivatives of the function, $f\left(x\right)=\sqrt[3]{x}$

$\begin{array}{rl}{f}^{\mathrm{\prime }}\left(x\right)& =\frac{d}{dx}\left[\sqrt[3]{x}\right]\\ & =\frac{1}{3}{x}^{-\frac{2}{3}}\end{array}$

So, at $x=1$

$\begin{array}{rl}{f}^{\mathrm{\prime }}\left(1\right)& =\frac{1}{3}{1}^{-\frac{2}{3}}\\ & =\frac{1}{3}\end{array}$

Step 4. Find f"(x)

$\begin{array}{rl}{f}^{\prime \prime }\left(x\right)& =\frac{d}{dx}\left[\frac{1}{3}{x}^{-\frac{2}{3}}\right]\\ & =\frac{1}{3}\frac{d}{dx}\left[x^{-\frac{2}{3}}\right]\\ & =\frac{1}{3}\cdot -\frac{2}{3}(x{)}^{-\frac{5}{3}}\\ & =-\frac{2}{9}{x}^{-\frac{5}{3}}\end{array}$

So at,$x=1$

$\begin{array}{rl}{f}^{\prime \prime }\left(1\right)& =-\frac{2}{9}(1{)}^{-\frac{5}{3}}\\ & =-\frac{2}{9}\end{array}$

Step 5. Find f'''(x)

$\begin{array}{rl}{f}^{\prime \prime \mathrm{\prime }}\left(x\right)& =\frac{d}{dx}[-\frac{2}{9}{x}^{-\frac{5}{3}}]\\ & =-\frac{2}{9}\frac{d}{dx}{x}^{-\frac{5}{3}}\\ & =-\frac{2}{9}\cdot -\frac{5}{3}{x}^{-\frac{8}{3}}\\ & =\frac{10}{27}{x}^{-\frac{8}{3}}\end{array}$

So, at $x=1$

$\begin{array}{rl}{f}^{\prime \prime \mathrm{\prime }}\left(1\right)& =\frac{10}{27}(1{)}^{-\frac{8}{3}}\\ & =\frac{10}{27}\end{array}$

Step 6. Find f''''(x)

$\begin{array}{rl}{f}^{\prime \prime \prime \prime }\left(x\right)& =\frac{d}{dx}\left[\frac{10}{27}{x}^{-\frac{8}{3}}\right]\\ & =\frac{10}{27}\cdot -\frac{8}{3}{x}^{-\frac{11}{3}}\\ & =-\frac{80}{81}{x}^{-\frac{11}{3}}\end{array}$

So, at $x=1$

$\begin{array}{rl}{f}^{\prime \prime \prime \prime }\left(1\right)& =-\frac{80}{81}(1{)}^{-\frac{11}{3}}\\ & =-\frac{80}{81}\end{array}$

Step 7. The fourth Taylor polynomial of the function 

Hence the fourth Taylor polynomial of the function $f\left(x\right)=\sqrt[3]{x}$at $x=1$

$\begin{array}{rl}{P}_4\left(x\right)& =1+\frac{1}{3}(x-1)+\frac{-\frac{2}{9}}{2!}(x-1{)}^2+\frac{\frac{10}{27}}{3!}(x-1{)}^3+\frac{-\frac{80}{81}}{4!}(x-1{)}^4\\ & =1+\frac{1}{3}(x-1)-\frac{1}{9}(x-1{)}^2+\frac{5}{81}(x-1{)}^3-\frac{10}{243}(x-1{)}^4\end{array}$