Answer
$\sqrt{2.1}-\sqrt{2}$ is larger than $\sqrt{9.1}-\sqrt{9}$
Work Step by Step
For $\sqrt{2.1}-\sqrt{2}$, consider $ f(x)=\sqrt{x}$, $a=2 $ and $\Delta x= 0.1 $, since
\begin{align*}
f'(x) &= \frac{1}{2\sqrt{x}}\\
f'( 2)&=\frac{1}{2\sqrt{2}}
\end{align*}
Then
\begin{align*}
\Delta f&=f(a+\Delta x)-f(a)\\
&\approx f'(a)\Delta f \\
&=\frac{0.1}{2\sqrt{2}} \\
&=0.03535
\end{align*}
For $\sqrt{9.1}-\sqrt{9}$, consider $ f(x)=\sqrt{x}$, $a=9 $ and $\Delta x= 0.1 $, since
\begin{align*}
f'(x) &= \frac{1}{2\sqrt{x}}\\
f'( 9)&=\frac{1}{6}
\end{align*}
Then
\begin{align*}
\Delta f&=f(a+\Delta x)-f(a)\\
&\approx f'(a)\Delta f \\
&=\frac{0.1}{6} \\
&=0.01666
\end{align*}
Hence $\sqrt{2.1}-\sqrt{2}$ is larger than $\sqrt{9.1}-\sqrt{9}$
