Answer:
[tex]t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3[/tex]
Step-by-step explanation:
We are given that
[tex]f(x)=7tan^{-1}(x)[/tex]
a=1
[tex]T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}[/tex]
Substitute n=3 and a=1
[tex]t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}[/tex]
[tex]f(x)=7tan^{-1}(x)[/tex]
[tex]f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}[/tex]
Where [tex]tan^{-1}(1)=\frac{\pi}{4}[/tex]
[tex]f'(x)=\frac{7}{1+x^2}[/tex]
Using the formula
[tex]\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}[/tex]
[tex]f'(1)=\frac{7}{2}[/tex]
[tex]f''(x)=\frac{-14x}{(1+x^2)^2}[/tex]
[tex]f''(1)=-\frac{7}{2}[/tex]
[tex]f''(x)=-14x(x^2+1)^{-2}[/tex]
[tex]f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})[/tex]
By using the formula
[tex](uv)'=u'v+v'u[/tex]
[tex]f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}[/tex]
[tex]f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}[/tex]
[tex]f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}[/tex]
Substitute the values
[tex]t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3[/tex]
[tex]t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3[/tex]
