Calculus of a Single Variable, Chapter 2, 2.1, Section 2.1, Problem 37

The given line is :-
x + 2y - 6 = 0
or, y = -(1/2)x + 3 (the line is represented in slope intercept form)
Thus, the slope of the line = -(1/2)
Now, the tangent to the curve f(x) = 1/(x^(1/2)) is parallel to the above line
Thus, the slope of the tangent = slope of the line = -(1/2).......(1)
The given function is:-
f(x) = 1/(x^(1/2))
differentiating both sides w.r.t 'x' we get
f'(x) = -(1/2)*[1/{x^(3/2)}]
Now, slope of the tangent = -(1/2)
Thus, -(1/2)*[1/{x^(3/2)}] = -(1/2)
or, x = 1
Putting the value of x =1 in the given equation of curve, we get
f(1) = y = 1
Hence the tangent passes through the point (1,1)
Thus, equation of the tangent at the point (1,1) and having slope = -(1/2) is :-
y - 1 = (-1/2)*(x - 1)
or, 2y - 2 = -x + 1
or, 2y + x - 3 = 0 is the equation of the tangent to the given curve at (1,1)