0

We have $$g(x) \equiv f(x,1) = \frac{1}{1-x}+1$$.

I would like to know whether $g(x)$ is increasing or decreasing. Re-writing $g(x) = \frac{2-x}{1-x}$, I obtain $$g'(x) = \frac{x-3}{(1-x)^{2}},$$ where $g'(x)=0$ at $x=3$. Then I obtain $$g''(x)=\frac{(1-x)^{2}-2(x-3)(1-x)}{(1-x)^{4}},$$ which at $x=3$ is positive. So I think $g$ is decreasing.

Also, $g$ is not differentiable at $x=1$ as $g$ is undefined at $x=1$. I don't know how to show it formally.

Are my answers correct?

OGC
  • 2,305

1 Answers1

3

You don't need the derivative to see if it's monotonic, as it a homographic function. Inded $1-x$ is decreasing, hence $\dfrac1{1-x}$ (as long as its sign is constant) is increasing, and $\dfrac1{1-x}+1$ is increasing too.

Your derivative is wrong, and it was simpler to derive $$\Bigl(\frac1{1-x}\Bigr)'=\frac1{(1-x)^2}>0.$$ As it is not defined at $x=1$, it is not differentiable thereat either.

Bernard
  • 175,478
  • What about the proof of $g$ being cts at $x=0$? – OGC Sep 13 '15 at 02:39
  • @OBC: Sorry, I forgot to mention that point, because at first I didn't grasp what you abbreviation (cts) meant. Unfortunately, you didn't check the condition $\Bigl\lvert\dfrac1{1-x}\Bigr\rvert$$<\dfrac12$ is valid in a neighbourhood of $0$, and indeed it is not, since the value of this absolute value at $x=0$ is $1$ and the function has to be proved to be continuous at $0$. I'll complete my answer in a moment. – Bernard Sep 13 '15 at 09:52
  • What value of $\delta$ did you pick to prove continuity? – OGC Sep 13 '15 at 18:29
  • I have seveal. The last and simplest one is $\frac\varepsilon2$. Otherwise $\dfrac\varepsilon{1+\varepsilon}$. – Bernard Sep 13 '15 at 18:33
  • Could you re-write your proof in $\epsilon$, $\delta$ format similar to mine. I cannot follow your answer. – OGC Sep 13 '15 at 18:37
  • Please remove your cty answer as I'm asking it on another post. – OGC Sep 13 '15 at 18:40
  • If you wish. My explanations are not clear enough? – Bernard Sep 13 '15 at 18:44
  • If you don't mind. Sorry. I only understand a step-by-step $\epsilon$, $\delta$ proof. Here's the latest post for $g$ cts at $x=0$ http://math.stackexchange.com/questions/1433919/prove-that-g-is-continuous-at-x-0. I hope people don't get confused now. – OGC Sep 13 '15 at 18:45