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.
Answer:
(4+√142)/(3) or x=(4-√142) /3
Step-by-step explanation:
(4+√142)/(3) or x=4-√142 /3
1. -6 ≤ x < -1 . . . . conjunction
2. x ≤ 6 . . or . . 10 ≤ x . . . . disjunction
3. 7 ≤ x < 12 . . . . conjunction
4. x < -9 . . or . . -3 ≤ x . . . . disjunction
5. 2 ≤ x ≤ 5 . . . . conjunction
6. x ≤ 54 . . or . . 66 ≤ x . . . . disjunction
7. 39 < x ≤ 43 . . . . conjunction
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Your problem statement provided no letters.
