Find the linearization of \( f(x)=\sqrt{x} \) at \(x=4 \) and then use it to find an approximation of \( \sqrt{5} \)

\( f^{\prime} (x)=\frac{1}{2\sqrt{x}} \)

\( f^{\prime} (4)=\frac{1}{2\sqrt{4}} = \frac{1}{4} \)

\( f (4)=\sqrt{4}=2 \)

\( L(x) = 2 + \frac{1}{4} (x-4)=\frac{1}{4}x+1 \)

\( \sqrt{5}=f(5) \approx L(5) = \frac{1}{4}(5)+1 = \frac{9}{4}=2.25 \)