The second degree Taylor polynomial of 1 x {\displaystyle {\frac {1}{x}}} at 4 is given by
f ( 4 ) + f ′ ( 4 ) ( x − 4 ) + f ″ ( 4 ) 2 ( x − 4 ) 2 {\displaystyle f(4)+f'(4)(x-4)+{\frac {f''(4)}{2}}(x-4)^{2}}
We already computed f ( 4 ) = 1 4 {\displaystyle f(4)={\frac {1}{4}}} and f ′ ( 4 ) = − 1 16 {\displaystyle f'(4)={\frac {-1}{16}}} . We have that f ″ ( x ) = 2 x 3 {\displaystyle f''(x)={\frac {2}{x^{3}}}} , so f ″ ( 4 ) = 2 64 = 1 32 {\displaystyle f''(4)={\frac {2}{64}}={\frac {1}{32}}} .
So the Taylor polynomial is
1 4 − 1 16 ( x − 4 ) + 1 64 ( x − 4 ) 2 {\displaystyle {\frac {1}{4}}-{\frac {1}{16}}(x-4)+{\frac {1}{64}}(x-4)^{2}} .
Answer: 1 4 − 1 16 ( x − 4 ) + 1 64 ( x − 4 ) 2 {\displaystyle \color {blue}{\frac {1}{4}}-{\frac {1}{16}}(x-4)+{\frac {1}{64}}(x-4)^{2}}
at 4 is given by
and 
. We have that 
, so 
.
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