<s>[INST] 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?
 [/INST] 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 $k=3$.
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