Answer
$y=-\frac{1}{2}x +\frac{3}{2} $
Work Step by Step
Begin by taking the derivative of $f(x)=\frac{1}{\sqrt x}$
$f'(x)= -\frac{1}{2\sqrt x^3} $
Find the equation of the given line
$x+2y-6=0$
$y=-\frac{1}{2}x+3$
Set the slope of the line equal to the tangent line slope
$ -\frac{1}{2\sqrt x^3} = -\frac{1}{2} $
Solve for x
$\frac{1}{\sqrt x^3} =1$,
$\sqrt[3] 1 = \sqrt[3] (\sqrt x^3)$,
$1=\sqrt x$
$1=x$
Plugging x back into the function to find a point
$f(1)=1$, $(1,1)$
Use point-slope to solve
$y-1=-\frac{1}{2}(x-1)$
$y=-\frac{1}{2}x + \frac{3}{2}$
This yields a line tangent to the graph and parallel to the given line.
