Calculus  Taylor
posted by COFFEE .
could you please help me with solving this problem?
#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!)(x1)^0 +(f(1)/1!)(x1)^1 + (f(1)/2!)(x1)^2
T2(x) = 2 + 1/4(x1) + 1/64(x1)^2
..do i solve for x???
which i tried..
0 = 2 + 1/4(x1) + 1/64(x1)^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(x1) + 1/64(x1)^2
..do i solve for x???"
The answer is T2(x) you just calculated.
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