Where does the function $f(x) = \frac{2x}{x - 7}$ have an increasing slope? Where does the function $f(x) = \frac{2x}{x - 7}$ have an increasing slope?


$a. x \le 0, x &gt; 7$
$b. x&lt;7$
$c. x &gt; 7$
$d. x \in \Bbb R, x \neq 7$


This question is from a test of mine in a pre-calculus course (so no calculus allowed in answering the question).I have no idea as to how to solve this problem. I can tell where the function is negative and where it is positive, but that's about it, and I'm fairly sure that's no use here.

Any ideas?
I'd go for $$\frac{2x}{x-7}=\frac{2(x-7)}{x-7} +\frac{14}{x-7}$$

The first term is constant, and the second one is just the graph of $1/x$ shifted and rescaled.
Differentiating gives f'(x) = -14/(x-7)^2 which is always negative. Drawing the curve of y = 2 + 14/(x-7) as suggested by newbie one gets a rectangular hyperbola with asymptotes x = 7, y = 2 showing negative gradient at all times.I'd be happier if the problem stated 'decreasing' rather than 'increasing' slope, in which case the answer would be d.

CORRECTION. I've just read the solution above, and must have misunderstood the phrase 'increasing slope' to mean 'positive slope". The slope is always negative, but changes as x changes, so can of course be increasing negatively.