QuestionFind the Taylor polynomial of degree 3 for the function $f(x)=\frac{x+7}{x-4}$ about $x=5$. $$T_{3}(x)=\square+\square(x-5)+\square(x-5)^{2}+\square(x-5)^{3}$$
Note: You can earn partial credit on this problem.

Studdy Solution
Construct the Taylor polynomial $T_3(x)$:
$$T_3(x) = f(5) + f'(5)(x-5) + \frac{f''(5)}{2!}(x-5)^2 + \frac{f'''(5)}{3!}(x-5)^3$$
Substitute the values:
$$T_3(x) = 12 - 11(x-5) + \frac{22}{2}(x-5)^2 + \frac{-66}{6}(x-5)^3$$
Simplify:
$$T_3(x) = 12 - 11(x-5) + 11(x-5)^2 - 11(x-5)^3$$
The Taylor polynomial of degree 3 is:
$$T_3(x) = 12 - 11(x-5) + 11(x-5)^2 - 11(x-5)^3$$