# 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.
$f\left(x\right)=\sqrt{x},a=4$
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brawnyN
$f\left(x\right)=\sqrt{x}$
$f\left(4\right)=\sqrt{4}=2$
${f}^{\prime }\left(x\right)=\frac{1}{2\sqrt{x}}$
${f}^{\prime }\left(4\right)=\frac{1}{2\sqrt{4}}=\frac{1}{4}$
The linearization L(x) of f at a=4:
L(x)=f(a)+f'(a)(x-a)
$L\left(x\right)=2+\frac{1}{4}\left(x-4\right)$
$L\left(x\right)=\frac{1}{4}x+1$
Results:
$L\left(x\right)=\frac{1}{4}x+1$