Calculus of a Single Variable, Chapter 2, 2.1, Section 2.1, Problem 38

Since the tangent line to function f(x) = 1 / sqrt(x -1) has to be parallel to the line x + 2y + 7 = 0, both needs to have the same slope. Equation of the straight line can be rewritten in slope-intercept form to : y = -x/2 - 7/2. Hence the desired slope of our line is -1/2.
And derivative of f(x) will provide the slope of the line tangent to f(x)d/dx(f(x)) = -1/2 * (x-1) ^ (-3/2)setting it equal to -1/2 -1/2 * (x-1) ^ (-3/2) = -1/2 or x = 2 provides one of the conditions
Corresponding y coordinate can be found from the f(x) = 1/sqrt(x - 1) equation, y = 1/sqrt(2 - 1) = 1The slope of the desired line is equal to -1/2, and we now have a point on the desired line (2, 1), as well the point of tangency
Using the point-slope, equation to desired line can be obtained as follows. Given slope m = -1/2 and a point on the desired line,(x0, y0), the equation of the line is given by(y - y0) = m(x - x0)y - 1 = -1/2(x - 2 ) y = -1/2x + 22y + x - 4 = 0