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.
Method A: 7.6666.... = 7 + 0.6666... = 7 + 2/3 = 21/3 + 2/3 = 23/3
Method B: 10(7.666...) - 1(7.666...) = 76.666... - 7.666... = 69.000...
(10 - 1)(7.666...) = 69
7.666... = 69/9 = 23/3
This expression is a perfect square trinomial. It has two answers but they don't have to be different. In this case it has to be the same - because it is a perfect square trinomial.
The additive inverse of something is whats added to the original to get 0
So in this case n - n = 0, which is B :)
Answer:
C. x =3
Step-by-step explanation:
Extraneous solution is that root of a transformed equation that doesn't satisfy the equation in it's original form because it was excluded from the domain of the original equation.
Let's solve the equation first
Hence, we can conclude that x=3 is an extraneous solution of the equation ..
