#### Answer

$x \in[1,2.9]$

#### Work Step by Step

Given $$f(x) =\sqrt{x}=x^{1/2},\ \ \ a=1 $$
Since
\begin{array}{rlrl}
f(x) & =x^{1 / 2} & & f(1)=1 \\
f^{\prime}(x) & =\frac{1}{2} x^{-1 / 2} & & f^{\prime}(1)=\frac{1}{2} \\
f^{\prime \prime}(x) & =-\frac{1}{4} x^{-3 / 2} & f^{\prime \prime}(1) & =-\frac{1}{4} \\
f^{\prime \prime \prime}(x) & =\frac{3}{8} x^{-5 / 2} & f^{\prime \prime \prime}(1) & =\frac{3}{8}
\end{array}
Then
\begin{aligned}
T_{3}(x) &=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3} \\
&= 1+\frac{1}{2}(x-1)-\frac{1}{4 \cdot 2 !}(x-1)^{2}+\frac{3}{8 \cdot 3 !}(x-1)^{3}\\
&=1+\frac{1}{2}(x-1)-\frac{1}{8}(x-1)^{2}+\frac{1}{16}(x-1)^{3}
\end{aligned}
Since
$$|f(x)-T_3(x)|= |\sqrt{x}-1-\frac{1}{2}(x-1)+\frac{1}{8}(x-1)^{2}-\frac{1}{16}(x-1)^{3}|$$
From the following graph, we have for $x \in[1,2.9]$ that the error does not exceed $0.25 .$ The error at $x=3$ appears to just exceed $0.25$