#### Answer

$y=\frac{1}{4}x+1$

#### Work Step by Step

The question asks us to find the equation of the tangent of the function
$f(x)=\sqrt x=x^{\frac{1}{2}}$ at the point (4,2)
Given that we are looking for the equation of a tangent, the first reasonable step would be to find the gradient at the points given, which can be found using the derivative function.
Using the power rule:
$f(x)=\sqrt x=x^{\frac{1}{2}}$
$f′(x)=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{\frac{-1}{2}}=\frac{1}{2\sqrt x}$
$f′(4)=\frac{1}{2\sqrt 4}=0.25$
Thus, the gradient of that function at the value x=4 is 0.25.
Using this information, we can find the intercept value of the tangent line.
The equation of the gradient is given by:
$\frac{y_{1}−y_{0}}{x_{1}−x_{0}}=m$
We can now use the points (4,2) and (0,c) (where c is the y-intercept).
$\frac{2−c}{4-0}=\frac{1}{4}$
Solving for c gives us c=1.
Thus, the equation of the tangent at (1,4) is $y=\frac{1}{4}x+1$