Find K such that the line is tangent to $y=-\frac{3}{4}x+4$ The function is $f(x)= \frac{k}{x}$ and the line is $y =-\frac{3}{4}x+3$
So the derivative of $f(x)$ should equal $-\frac{3}{4}$
If I take $k$ to be a constant I can remove it from the fraction like so 
$$f'(x)=k \frac{d}{dx} \frac{1}{x}$$
to get $-\frac{k}{x^2}$ but then I try to solve for $k$
$$-\frac{k}{x^2}=-\frac{3}{4}$$ 
and this is where I get stuck. So should I not even remove $k$ from the differentiation or have I gone in a completely wrong direction?
 A: From the given equation we obtain
$$-\frac{3}{4}x+3=\frac{k}{x}$$ so we get
$$-\frac{3}{4}x^2+3x-k=0$$ solving for $x$ we get
$$x=2\pm \frac{2}{3}\sqrt{9-3k}$$ since the line is a tangent line, we get
$$9-3k=0$$
A: If the tangent occurs at $x = t$, we have $\displaystyle\frac{k}{t^2} = \frac{3}{4}$, as you've already found, but we also must have $\displaystyle\frac{k}{t} = -\frac{3}{4}t + 3$. 
So we obtain $\displaystyle k = -\frac{3t^2}{4} + 3t$, and subbing into the first equation gives $\displaystyle -\frac{3}{4} + \frac{3}{t} = \frac{3}{4}$, from which we obtain $\displaystyle \frac{1}{t} = \frac{1}{2}$ and finally $t=2$. So $\displaystyle \frac{k}{t^2} = \frac{3}{4}$ now becomes $\displaystyle\frac{k}{4} = \frac{3}{4}$, and so $k=3$.
A: Note that$$-\frac k{x^2}=-\frac34\iff x=\frac2{\sqrt3}\sqrt k.$$Now, consider the value of $y$ when $x=\frac2{\sqrt3}\sqrt k$. And consider the value of $-\frac34x+3$ when $x=\frac2{\sqrt3}\sqrt k$. They are equal when (and only when) $k=3$.

A: Now, $$k=\frac{3x^2}{4},$$ which gives $$y=\frac{3x}{4},$$
$$\frac{3x}{4}=-\frac{3}{4}x+3$$ and $$x=2.$$
Can you end it now? 
A: If $xy = k$, then $y + xy' = 0$ and $ky' + y^2 = 0$, and the tangent $y = y'_0(x - x_0)+y_0 = y'_0x + 2y_0$. Thereby $$k = -\frac{y^2_0}{y'_0} = 3$$ where " the line is $-\frac{3}{4}x + 3$".
A: The question should be 

Find $k$ such that the line $y=-\frac34x+3$ is tangent to the curve $f(x)=\frac{k}{x}$.

More systematic way!

The tangent line to $y=f(x)$ at the point $(x_0,f(x_0))$ is:
  $$y=f(x_0)+f'(x_0)(x-x_0)=\underbrace{f'(x_0)}_{slope}x+\underbrace{f(x_0)-f'(x_0)x_0}_{intercept}$$

You can make up the system of two equations:
$$\begin{cases}f'(x_0)=-\frac34 \\ f(x_0)-f'(x_0)x_0=3\end{cases}\Rightarrow \begin{cases}-\frac{k}{x_0^2}=-\frac34\\ \frac{k}{x_0}+\frac{k}{x_0^2}\cdot x_0=3\end{cases}\Rightarrow \begin{cases}k=\frac34x_0^2\\ x_0=\frac23k,k\ne 0\end{cases}\Rightarrow \begin{cases}k=3\\ x_0=2\end{cases}.$$
