Answer
$k = 3$
Work Step by Step
To solve this problem, we need to identify point $(x,y)$ where both the function and the equation of the tangent line intersect. To do this, we can equate both to produce the following: $$\frac{k}{x} = -\frac{3}{4}x + 3$$ By rewriting the new equation such that it is equal to zero, we achieve the following: $$0 = -\frac{3}{4}x^{2} + 3x - k$$ We can now solve for x by using the quadratic formula: $$x = \frac{-b +/- (b^{2} - 4ac)^{\frac{1}{2}} }{2a}$$ However, an interesting corollary to the quadratic formula states that the expression inside the square root, $b^{2} - 4ac$, is positive when $x$ has 2 solutions, $0$ when it only has 1 solution, and negative when $x$ has no solution in real numbers. Since we know that, by definition, the tangent line only intersects at one point $(x,y)$ with the function $f(x)$, we know that $x$ can only have one solution and, as such: $$b^{2} - 4ac = 0$$ Plug in the appropriate values: $$(3)^{2} - 4(-\frac{3}{4})(-k) = 0$$ $$9-3k = 0$$ Therefore: $$k = 3$$
