Question

Find the first three nonzero terms of the Taylor expansion for 2/x expanded about the value 3

A)2/3-2/9(X-3)+4/27(X-3)²+........
B)-2/3+2/9(X-3)+(-2/27)(X-3)²+........
C)2/3-2/9(X-3)+2/27(X-3)²+........
D)-2/3-2/9(X-3)+(-4/27)(X-3)²+........

Answer 1

The first three nonzero terms of the Taylor expansion for 2/x expanded about the value 3 are 2/3 - 2/9(x-3) + 4/27(x-3)².

Step-by-step explanation:

The Taylor expansion for a function f(x) about a value a is given by the formula:

`f(x) = f(a) + f'(a)(x-a) + (f''(a))/(2!)(x-a)^2 + (f'''(a))/(3!)(x-a)^3 + ...`

In this case, the function is
`f(x) = (2)/(x)`a is 3.

To find the Taylor expansion, we need to find the derivatives of f(x) and evaluate them at x = 3.

The first derivative of f(x) is
`f'(x) = -(2)/(x^2).`

When x = 3,
`f'(3) = -(2)/(9).`

The second derivative of f(x) is
`f''(x) = (4)/(x^3).`

When x = 3,
`f''(3) = (4)/(27).`

Substituting these values into the Taylor expansion formula:

`f(x) = f(3) + f'(3)(x-3) + (f''(3))/(2!)(x-3)^2 + ...`

The first three nonzero terms of the expansion are:

`(2)/(3) - (2)/(9)(x-3) + (4)/(27)(x-3)^2 + ...`