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?


*

*$a. x \le 0, x > 7$

*$b. x<7$

*$c. x > 7$

*$d. x \in \Bbb R, x \neq 7$


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?
 A: 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.
A: $$\frac{2x}{x-7} = \frac{2x-14+14}{x-7} =2 + \frac{14}{x-7}.$$
Are you allowed to use basic properties of $\frac{1}{x}$?
A: 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.
A: $f(x)$ has an increasing slope when the second derivative of $f(x)$ with respect to $x$ is positive. The second derivative is $28/(x-7)^3$, which is $>0$ when $x>7$, $<0$ when $x<7$, and undefined when $x=7$. Therefore, the answer is $c (x>7)$.
Alternately, you could graph your function, draw some tangent lines to the curve at different points, and just notice where the slopes of those tangent lines are increasing. This way you won't have to do any differentiation. Hope this helps.
