Which answer describes the transformation of f (x) = x^2 −1 to g(x)=(x+2)^2−1 ?
1 answer:
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Answer:
The Taylor series of f(x) around the point a, can be written as:
Here we have:
f(x) = 4*cos(x)
a = 7*pi
then, let's calculate each part:
f(a) = 4*cos(7*pi) = -4
df/dx = -4*sin(x)
(df/dx)(a) = -4*sin(7*pi) = 0
(d^2f)/(dx^2) = -4*cos(x)
(d^2f)/(dx^2)(a) = -4*cos(7*pi) = 4
Here we already can see two things:
the odd derivatives will have a sin(x) function that is zero when evaluated in x = 7*pi, and we also can see that the sign will alternate between consecutive terms.
so we only will work with the even powers of the series:
f(x) = -4 + (1/2!)*4*(x - 7*pi)^2 - (1/4!)*4*(x - 7*pi)^4 + ....
So we can write it as:
f(x) = ∑fₙ
Such that the n-th term can written as: