Question

a graphing calculator is recommended.
find the taylor polynomial ( t _ { 3 } ( x ) ) for the function ( f ) centered at the number ( a ).
( f ( x ) = 5 \tan ^ { - 1 } ( x ), quad a = 1 )
( t _ { 3 } ( x ) = )
graph ( f ) and ( t _ { 3 } ) on the same screen.

Explanation:

Step1: Recall the Taylor polynomial formula

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

Step2: Find the first - order derivative

Given \(f(x)=5\tan^{- 1}(x)\), use the formula \((\tan^{-1}(x))^{\prime}=\frac{1}{1 + x^{2}}\). Then \(f^{\prime}(x)=\frac{5}{1+x^{2}}\)

Step3: Find the second - order derivative

Use the quotient rule \((\frac{u}{v})^{\prime}=\frac{u^{\prime}v - uv^{\prime}}{v^{2}}\), where \(u = 5\) and \(v=1 + x^{2}\). So \(f^{\prime\prime}(x)=\frac{0\times(1 + x^{2})-5\times2x}{(1 + x^{2})^{2}}=\frac{-10x}{(1 + x^{2})^{2}}\)

Step4: Find the third - order derivative

Use the quotient rule again. Let \(u=-10x\) and \(v=(1 + x^{2})^{2}\). First, \(u^{\prime}=-10\) and \(v^{\prime}=2(1 + x^{2})\times2x = 4x(1 + x^{2})\). Then \(f^{(3)}(x)=\frac{-10\times(1 + x^{2})^{2}-(-10x)\times4x(1 + x^{2})}{(1 + x^{2})^{4}}=\frac{-10(1 + x^{2})+40x^{2}}{(1 + x^{2})^{3}}=\frac{-10 + 30x^{2}}{(1 + x^{2})^{3}}\)

Step5: Evaluate \(f(x)\), \(f^{\prime}(x)\), \(f^{\prime\prime}(x)\) and \(f^{(3)}(x)\) at \(a = 1\)

• \(f(1)=5\tan^{-1}(1)=\frac{5\pi}{4}\)
• \(f^{\prime}(1)=\frac{5}{1 + 1^{2}}=\frac{5}{2}\)
• \(f^{\prime\prime}(1)=\frac{-10\times1}{(1 + 1^{2})^{2}}=-\frac{5}{2}\)
• \(f^{(3)}(1)=\frac{-10+30\times1^{2}}{(1 + 1^{2})^{3}}=\frac{20}{8}=\frac{5}{2}\)

Step6: Construct \(T_{3}(x)\)

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Answer:

\(T_{3}(x)=\frac{5\pi}{4}+\frac{5}{2}(x - 1)-\frac{5}{4}(x - 1)^{2}+\frac{5}{12}(x - 1)^{3}\)