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## Solution

Your input: find the linear approximation to $f(x)=\sqrt{x}$ at $x_0=4$.

A linear approximation is given by $L(x)\approx f(x_0)+f^{\prime}(x_0)(x-x_0)$.

We are given that $x_0=4$.

Firstly, find the value of the function at the given point: $y_0=f(x_0)=2$.

Secondly, find the derivative of the function, evaluated at the point: $f^{\prime}\left(4\right)$.

Find the derivative: $f^{\prime}\left(x\right)=\frac{1}{2 \sqrt{x}}$ (steps can be seen here).

Next, evaluate the derivative at the given point to find slope.

$f^{\prime}\left(4\right)=\frac{1}{4}$.

Plugging the values found, we get that $L(x)\approx 2+\frac{1}{4}\left(x-\left(4\right)\right)$.

Or, more simply: $L(x)\approx \frac{1}{4} x+1$.

Answer: $L(x)\approx \frac{1}{4} x+1$.