# Calculus - Taylor

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#1) Find the Taylor polynomial Tn(x) for the function 'f' at the number 'a'.

f(x) = sqrt(3+x^2) ; a=1; n=2;

my work so far:

f (x) = sqrt(3+x^2) = (3+x^2)^(1/2)
f ' (x) = (1/2)(3+x^2)^(-1/2)
f '' (x) = (-1/4)(3+x^2)^(-3/2)
f ''' (x) = (3/8)(3+x^2)^(-5/2)

f (x) = 2
f ' (x) = 1/4
f '' (x) = -1/32
f ''' (x) = 3/256

T2(x) = (f(1)/0!)(x-1)^0 +(f(1)/1!)(x-1)^1 + (f(1)/2!)(x-1)^2
T2(x) = 2 + 1/4(x-1) + -1/64(x-1)^2
..do i solve for x???

which i tried..

0 = 2 + 1/4(x-1) + -1/64(x-1)^2
2 = 1/4x - 1/4 - 1/64x^2 + 1/32x - 1/64
8 = x - 1 - 1/16x^2 + 1/8x - 1/16
64 = 8x - 8 - 1/2x^2 + 1/8x - 1/16
72.5 = 9x - 1/2x^2
145 = 18x - 1x^2
-145 = -18x + x^2
145/18 = x + x^2
.. and im not sure where to go with this..

"T2(x) = 2 + 1/4(x-1) + -1/64(x-1)^2
..do i solve for x???"

The answer is T2(x) you just calculated.

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