How do you do this question?

Step-by-step explanation:

The Taylor series expansion is:

Tₙ(x) = ∑ f⁽ⁿ⁾(a) (x − a)ⁿ / n!

f(x) = 1/x, a = 4, and n = 3.

First, find the derivatives.

f⁽⁰⁾(4) = 1/4

f⁽¹⁾(4) = -1/(4)² = -1/16

f⁽²⁾(4) = 2/(4)³ = 1/32

f⁽³⁾(4) = -6/(4)⁴ = -3/128

Therefore:

T₃(x) = 1/4 (x − 4)⁰ / 0! − 1/16 (x − 4)¹ / 1! + 1/32 (x − 4)² / 2! − 3/128 (x − 4)³ / 3!

T₃(x) = 1/4 − 1/16 (x − 4) + 1/64 (x − 4)² − 1/256 (x − 4)³

f(x) = 1/x has a vertical asymptote at x=0 and a horizontal asymptote at y=0.  So we can eliminate the top left option.  That leaves the other three options, where f(x) is the blue line.

Now we have to determine which green line is T₃(x).  The simplest way is to notice that f(x) and T₃(x) intersect at x=4 (which makes sense, since T₃(x) is the Taylor series centered at x=4).

The bottom right graph is the only correct option.

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