# Suppose f(2) = 1, f'(2) = 1, f"(2) = 0,

Suppose f(2) = 1, f'(2) = 1, f"(2) = 0, and f(3)(2) = 12.
Find the third-order Taylor polynomial for f centered at 2 and use this polynomial to estimate f(1.9).
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Debbie Moore
Step 1:Concept
Taylor polynomial is
$f\left(x\right)=f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)+\frac{f{}^{″}\left(a\right){\left(x-a\right)}^{2}}{2!}+\frac{f{}^{‴}\left(a\right){\left(x-a\right)}^{3}}{3!}+\dots$
Step 2:Solution
Plugging the values at a=2
$f\left(x\right)=f\left(2\right)+{f}^{\prime }\left(2\right)\left(x-2\right)+\frac{f{}^{″}\left(2\right){\left(x-2\right)}^{2}}{2!}+\frac{f{}^{‴}\left(2\right){\left(x-2\right)}^{3}}{3!}$
$=1+1\left(x-2\right)+\frac{0{\left(x-2\right)}^{2}}{2}+\frac{12{\left(x-2\right)}^{3}}{3!}$
$=1+x-2+2{\left(x-2\right)}^{3}$ At x=1.9
$=1+1.9-2+2{\left(1.9-2\right)}^{3}f\left(1.9\right)=0.898$
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eskalopit
Given,
$f\left(2\right)=1,{f}^{\prime }\left(2\right)=1,f{}^{″}\left(2\right)=0,{f}^{3}\left(2\right)=12$
$f\left(n\right)=f\left(a\right)+{f}^{\prime }\left(a\right)\left(n-a\right)+f{}^{″}\left(a\right)\frac{{\left(n-a\right)}^{2}}{2!}+f{}^{‴}\left(a\right)\frac{{\left(n-a\right)}^{3}}{3!}$
$f\left(n\right)=f\left(2\right)+{f}^{\prime }\left(2\right)\left(n-2\right)+f{}^{″}\left(2\right)\frac{{\left(n-2\right)}^{2}}{2!}+f{}^{‴}\left(2\right)\frac{{\left(n-a\right)}^{3}}{3!}$
$=1+1\left(n-2\right)+\left(0\right)\frac{{\left(n-2\right)}^{2}}{2}+12\frac{{\left(n-2\right)}^{3}}{6}$
$=1+n-2+0+2{\left(n-2\right)}^{3}$
$f\left(n\right)=2{\left(n-2\right)}^{3}+n-1$
$f\left(1.9\right)=2{\left(1.9-2\right)}^{3}+1.9-1$
$=2{\left(-0.1\right)}^{3}+0\cdot 9=-2\left(0.001\right)+0.9$
=0.9-0.002
=0.878
$\therefore f\left(1.9\right)=0.878$