\(\displaystyle{f{{\left({x}\right)}}}=\sqrt{{{x}}}\)

\(\displaystyle{f{{\left({4}\right)}}}=\sqrt{{{4}}}={2}\)

\(\displaystyle{f}'{\left({x}\right)}={\frac{{{1}}}{{{2}\sqrt{{{x}}}}}}\)

\(\displaystyle{f}'{\left({4}\right)}={\frac{{{1}}}{{{2}\sqrt{{{4}}}}}}={\frac{{{1}}}{{{4}}}}\)

The linearization L(x) of f at a=4:

L(x)=f(a)+f'(a)(x-a)

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

\(\displaystyle{L}{\left({x}\right)}={\frac{{{1}}}{{{4}}}}{x}+{1}\)

Results:

\(\displaystyle{L}{\left({x}\right)}={\frac{{{1}}}{{{4}}}}{x}+{1}\)

\(\displaystyle{f{{\left({4}\right)}}}=\sqrt{{{4}}}={2}\)

\(\displaystyle{f}'{\left({x}\right)}={\frac{{{1}}}{{{2}\sqrt{{{x}}}}}}\)

\(\displaystyle{f}'{\left({4}\right)}={\frac{{{1}}}{{{2}\sqrt{{{4}}}}}}={\frac{{{1}}}{{{4}}}}\)

The linearization L(x) of f at a=4:

L(x)=f(a)+f'(a)(x-a)

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

\(\displaystyle{L}{\left({x}\right)}={\frac{{{1}}}{{{4}}}}{x}+{1}\)

Results:

\(\displaystyle{L}{\left({x}\right)}={\frac{{{1}}}{{{4}}}}{x}+{1}\)