Answer:
The generated Taylor series is;
f(x) = 1/6 + (x-3)/36 + (x-3)^2/216 + (x-3)^3/1296......
Kindly note that the series here is truncated at the third derivative, but if wanted, evaluation could be more than this
Step-by-step explanation:
Here, we are to find the Taylor series generated by f at x = a
Mathematically;
f(x)= 1/(9-x)
The f(a) at this point is calculated below;
f(a) = 1/(9-3) = 1/6
The first derivative is calculated below;
f'(x) = 1/(9-x)^2
f'(3) = 1/6^2 = 36
The second derivative is calculated below;
f''(x) = 2/(9-x)^3
f''(3) = 2/(9-3)^3 = 2/6^3 = 2/216 = 1/108
The third derivative is calculated as follows
f'''(x) = 6/(9-x)^4
f'''(3) = 6/(9-3)^4 = 6/6^4 = 1/216..
The Taylor series is thus given by;
f(x) = f(a) + f'(a) (x-a) + f''(a) (x-a)^2/2! + f'''(a) (x-a)^3/3! +...
Substituting the values of 3 for a at each of the points , we obtain the Taylor series as shown below;
f(x) = 1/6 + (x-3)/36 + (x-3)^2*1/108*1/2 + (x-3)^3*1/216*1/6 +...
f(x) = 1/6 + (x-3)/36 + (x-3)^2/216 + (x-3)^3/1296......
