# Small proof on monotonic functions

1. Feb 23, 2012

### Bipolarity

I actually made this question up while studying some chemistry. The problem is easy to visualize, but I'm trying to formalize to help myself think more rigorously. To be precise I sort of thought about how you could prove that a reduction in vapor pressure causes a depression freezing point in an ideal solution.

Suppose that $f(x)$ and $g(x)$ are both real-valued differentiable functions defined for all x.

It is known that:
$f'(x)>0$ for all x
$g'(x)>0$ for all x
There exists exactly one value of $c$ such that $f(c) = g(c)$
There exists a $d$ such that $f(d) = g(d)-5$

Prove that $d<c$

I will be very thankful if someone could help me out. Again this is a problem I made up from my studies in chemistry. Not a homework problem.

Thanks!

BiP

Last edited: Feb 23, 2012
2. Feb 23, 2012

### micromass

I don't think it's true.

Take f(x)=2x and g(x)=x+1. Then f(0)=g(0)=1, but there certainly exists a point d>0 such that f(d)=g(d)+5. Take d approximately 3.20194.

3. Feb 23, 2012

### Bipolarity

Ah my mistake. I forgot to clarify that there is exactly one point of intersection between
f(x) and g(x).

Sorry!

BiP

4. Feb 23, 2012

### micromass

Still not true. Take f(x)=ex and g(x)=x+1.

5. Feb 23, 2012

### Bacle2

Just to throw an idea:

Define h(x): g(x)-f(x)

Then: h(c)=g(c)-f(c)=0

h(d)=g(d)-f(d)=5

And you know h(x) is differentiable . Then, using the MVT, there is an x in (c,d) with :

h'(x)=[h(d)-h(c)]/(d-c)=5/(d-c)

Then h'(x) --the difference (g-f) -- is positive when d>c, and otherwise negative, and

you know this difference is 0 at one point, i.e., at c.

So you're saying that the only choice is when one function grows faster than the other before- or after- they meet.

Last edited: Feb 23, 2012
6. Feb 23, 2012

### Bipolarity

Bacle2, I guess you've solved it. But micro's counterexamples were also correct, so I smell a paradox! Is giving me the goosebumps!

BiP

7. Feb 23, 2012

### Bacle2

Actually, I think my layout suggests that your hypothesis needs (at least tweaking), and
your conditions need to be strengthened, so at least from that I don't think there is a paradox.

8. Feb 23, 2012

### micromass

Yes, if you add the hypothesis that

$$g^\prime(x)>f^\prime(x)>0$$

for all x, then what you say is true.

9. Feb 23, 2012

### Bipolarity

Wait, micro, could you please walk me through how adding that condition consolidates Bacle's proof? I can understand the intuition behind it but I can't see it's role in the proof itself.

Also, I think you mean $$f^\prime(x)>g^\prime(x)>0$$

BiP

10. Feb 23, 2012

### micromass

Yes, I meant it the other way around.

Let's use Bacle's idea and introduce the function h=f-g.

We know that h(c)=0 and h(d)=-5 and h'(x)>0 for all x.

This also shows that the assumption that f' and g' is >0 is not necessary.

Assume that d>c. Using the mean-value theorem, we know that there is a b between c and d such that

$$h^\prime(b)=\frac{h(d)-h(c)}{d-c}$$

but the left-hand side is postive. The right-hand side evaluates to

$$\frac{-5}{d-c}$$

and is negative. This is a contradiction. So our assumption that d>c is false.

11. Feb 23, 2012

### Bipolarity

Amazing! Thanks! How do you guys do these so fast and elegantly?

BiP

12. Feb 23, 2012

### lugita15

Bipolarity, how exactly is this related to the chemistry problem you were discussing?

13. Feb 23, 2012

### Bipolarity

I'm glad you asked. When a solute is dissolved in a pure solvent and forms an ideal solution, the vapor pressure pressure of the solvent drops. (Raoult's law)

If you look at the phase diagram of any solvent, the added solvent essentially shifts the vapor pressure curve downward (this is only an approximation!) and it intersects with the solid-gas curve at a lower temperature, resulting in a lower triple point.

In my problem, f(x) was essentially the solid-gas curve. g(x) was the vapor pressure (or liquid-gas) curve, and the downward shift was done to prove that the curves now intersect at a lower temperature, hence the triple-point temperature depression.

It was difficult for me to initially realize that the solid-gas curve has a higher slope than the liquid-gas curve if you extrapolate it, which explains the initial confusion in this thread. Calculus be thanked!

Refer to this picture:

Pretty obvious from the diagram, it's just that I'm a stickler for abstract rigor. Dunno if that's good or bad thing.

BiP

Last edited: Feb 23, 2012