This is from MIT OCW single variable calculus, in a section about learning to sketch curves. The material explains that if f'(x) is negative then f(x) is decreasing - this makes sense to me geometrically, if the slope of the tangent is negative then the graph is obviously decreasing.

Now for the question:

Sketch the graph of y = x/(x+4); find the intervals on which it is increasing and decreasing and decide how many solutions there are to y = 0."

And the solution:

y = x/(x+4), y' = 4/(x+4)^2

Increasing on: -4 < x < infinity

Decreasing on: -infinity < x < -4

I can see that the solution is correct, but as far as I can tell, y' > 0 for all values of x (except x = -4).

Is there something I'm missing, algebraically? Or am I just supposed to look at the graph and infer that this is some special case of y'?

  • 1
    $\begingroup$ The function is increasing at every point. Why do you say it is decreasing in $(- \infty , -4)$? This is wrong. $\endgroup$ – Crostul Nov 13 '16 at 23:07
  • $\begingroup$ The function decreases when x < -4 E.g. -5 / (-5 + 4) = -5 / -1 = 5; don't negative numbers divided by negative numbers become positive? $\endgroup$ – Xianny Nov 13 '16 at 23:08
  • $\begingroup$ I don't get your point. You know that $f(-5)=5$, so what? $\endgroup$ – Crostul Nov 13 '16 at 23:09
  • $\begingroup$ @Xianny That´s not true. $\endgroup$ – callculus Nov 13 '16 at 23:09
  • 1
    $\begingroup$ The function is strictly increasing on any interval where it is defined. This does not mean that it is increasing on $(-5,-3)$ for example, since that interval includes point $-4$ where the function is not defined. $\endgroup$ – dxiv Nov 13 '16 at 23:12

enter image description here

The plot shows that the function does increase for every value of $x$.

As your derivative test did.

There is no decreasing behaviour.

  • 1
    $\begingroup$ The function is strictly increasing on $(-\infty,-4)$ and $(-4,\infty)$, but it's not monotonic on $\mathbb{R} \setminus \{-4\}$. I think that might be part of OP's confusion. $\endgroup$ – dxiv Nov 13 '16 at 23:23
  • 1
    $\begingroup$ @dxiv You're surely right about the monotony. It seemed me the OP had some issues also with a ghost decreasing part, so I plotted the function to show him at least the trend! $\endgroup$ – Von Neumann Nov 13 '16 at 23:24
  • $\begingroup$ thanks - getting a computer plotted graph helps me confirm the answer sheet is either wrong or I'm misunderstanding it! that's definitely where the confusion is coming from. $\endgroup$ – Xianny Nov 13 '16 at 23:25
  • $\begingroup$ @Xianny The function does increase for every value of $x$. At $x = -4$ the function is not defined, since the limit $4^+$ and $4^-$ are different. So you can tell the function does increase but not monotonically as dxiv suggested. If you have other doubts, please write them! $\endgroup$ – Von Neumann Nov 13 '16 at 23:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.