# Find the linearization L(x) of the function at a. f(x)=\sqrt{x}, a=4

Find the linearization L(x) of the function at a.
$$\displaystyle{f{{\left({x}\right)}}}=\sqrt{{{x}}},{a}={4}$$

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brawnyN
$$\displaystyle{f{{\left({x}\right)}}}=\sqrt{{{x}}}$$
$$\displaystyle{f{{\left({4}\right)}}}=\sqrt{{{4}}}={2}$$
$$\displaystyle{f}'{\left({x}\right)}={\frac{{{1}}}{{{2}\sqrt{{{x}}}}}}$$
$$\displaystyle{f}'{\left({4}\right)}={\frac{{{1}}}{{{2}\sqrt{{{4}}}}}}={\frac{{{1}}}{{{4}}}}$$
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:
$$\displaystyle{L}{\left({x}\right)}={\frac{{{1}}}{{{4}}}}{x}+{1}$$
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