Answer
$x+2y-4=0$
Work Step by Step
The line has an equation:
$x+2y+7=0$
$2y=-x-7$
$y=-\frac{1}{2}x-\frac{7}{2}$
The slope of the line is $-\frac{1}{2}$. For a line to be parallel that is tangent to $f$, the tangent should also have a slope of $-\frac{1}{2}$.
$f'(x)=-\frac{1}{2}$
$f(x)=\frac{1}{\sqrt{x-1}}$
$f(x)=(x-1)^{-\frac{1}{2}}$
$f'(x)=(-\frac{1}{2})(x-1)^{(-\frac{1}{2}-1)}(1)$
$f'(x)=-\frac{1}{2}(x-1)^{-\frac{3}{2}}$
$f'(x)=-\frac{1}{2(\sqrt{x-1})^3}$
$f'(x)=-\frac{1}{2}$
$-\frac{1}{2}=-\frac{1}{2(\sqrt{x-1})^3}$
$(\sqrt{x-1})^3=1$
$x=2$
$f(2)=\frac{1}{\sqrt{(2)-1}}$
$f(2)=1$
The line that is tangent to $f$ and is parallel to $x+2y+7=0$ has a slope of $-\frac{1}{2}$ and passes through the point $(2,1)$.
$y=mx+c$
$(1)=(-\frac{1}{2})(2)+c$
$c=2$
Therefore, the line has the equation,
$y=-\frac{1}{2}x+2$
$2y=-x+4$
$x+2y-4=0$
