Anyone mind explaining how to find the second-degree Taylor polynomial for f(x)=4x^2-7x+2 about x=0? Even just some guidance would be appreciated.
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f(x) = 4x² - 7x + 2
f '(x) = 8x - 7
f "(x) = 8
f(0) = 2
f '(0) = -7
f "(0) = 8
The formula for each "n" term is:
f^(n) (a) * (x-a)^n
-----------------------
n!,
where f^(n) (a) is the "nth" derivative of f(a), and
a is the number the poly is being centered about.
In this case, a=0.
P2(x) = (0(x-0)^0)/0! + (-7)(x-0)/1! + 8(x-0)²/2!
= 0 - 7x + 8x²/2
= -7x + 4x²
So basically, you start off by finding the derivatives necessary to obtain the poly-degree the question is asking for. Then you evaluate those derivatives for the value of x that the poly is centered about, then follow that formula above for each term, starting with n=0
f '(x) = 8x - 7
f "(x) = 8
f(0) = 2
f '(0) = -7
f "(0) = 8
The formula for each "n" term is:
f^(n) (a) * (x-a)^n
-----------------------
n!,
where f^(n) (a) is the "nth" derivative of f(a), and
a is the number the poly is being centered about.
In this case, a=0.
P2(x) = (0(x-0)^0)/0! + (-7)(x-0)/1! + 8(x-0)²/2!
= 0 - 7x + 8x²/2
= -7x + 4x²
So basically, you start off by finding the derivatives necessary to obtain the poly-degree the question is asking for. Then you evaluate those derivatives for the value of x that the poly is centered about, then follow that formula above for each term, starting with n=0