Find the taylor series generated by f at x = a. f(x) = 1/9-x, a = 3.

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

a

step-by-step explanation:

(0,,,0)

step-by-step explanation:

the table of x and y values is shown in the attached pic(part of question)

we have to decide according that graph

given points are (0, 0) (–1, 0), (2, 0)

when we graph these points on graph they show an increasing line as shown in the attached pic