Consider the function $f(x) = \sqrt{x+1}$.
Let $T_n$ be the $n^{th}$ degree Taylor approximation of $f(10)$ about $x = 8$.

Find: $T_{1} =$ $T_{2} =$
Use 3 decimal places in your answer, but make sure you carry all decimals when performing calculations

$T_{2}$ is an (over/under) estimate of $f(10)$.

If $R_{2}$ is the remainder given by the Lagrange Remainder Formula:
$\left|R_{2}\right| \le$