Respuesta :

LammettHash
Assuming the function is [tex]f(x)=\dfrac{5x-5}{x+1}[/tex], and not [tex]5x-5x+1=1[/tex] which would be its own Maclaurin polynomial right away.

[tex]\dfrac{5x-5}{x+1}=\dfrac{5(x-1)}{x+1}=\dfrac{5(x+1-2)}{x+1}=5-\dfrac{10}{1+x}[/tex]

Recall the geometric power series,

[tex]\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}[/tex]

where [tex]|x|<1[/tex]. Replace [tex]x\mapsto-x[/tex] and we have

[tex]\displaystyle\sum_{n=0}^\infty (-x)^n=\frac1{1+x}[/tex]

and so

[tex]f(x)=5-10\displaystyle\sum_{n=0}^\infty(-x)^n[/tex]

We want the 4th degree polynomial, so

[tex]p_4(x)=-5+10x-10x^2+10x^3-10x^4[/tex]

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