Applied maths, monotonic function

[nobahar]

Hello,

The equation is from a chemistry calculation; the textbook claims that the function is monotonic, without specifying whether it is monotonically increasing or decreasing.

Depending on the starting conditions, the function can look different; I basically want to know if the following is correct, and if it is a good general approach for these types of problems - that is, determining if the function in monotonically increasing or decreasing, or not monotonic.

So, the equation I have is as follows, $C_1$, $C_2$, etc are constants:
f(x) \ = \ C_4 \ \left( {\frac{C_1 \ C_2 \ + \ C_1 \ x}{x^2 \ + \ x \ C_1 \ + C_1 \ C_2}} \right) \ - \ x \ + \ {\frac{C_3}{x}}

I figured that if the function is monotonically increasing, then $f'(x) >= 0$, and if it is decreasing, then $f'(x) <= 0$. So I (think) I took the derivative, and got:
f&#039;(x) \ = \ C_4 \ \left( {\frac{-2 \ C_1 \ C_2 \ x \ - \ C_1 \ x^2}{(x^2 \ + \ C_1 \ x \ + \ C_1 \ C_2)^2}} \right) \ - \ 1 \ - \ {\frac{C_3}{x^2}}

Since this is applied, I believe there are some constraints, as follows: $x \ >= \ 0$, $C_1 \ > \ 0$, $C_2 \ > \ 0$, $C_3 \ > \ 0$, and $C_4 \ > \ 0$. With these in mind, the denominator of the first fraction is always > 0 (its a summation of positive numbers and is squared anyway); focusing on the numerator of the same fraction, $2 \ C_1 \ C_2 \ x > \ 0$, and $C_1 \ x \ > \ 0$, since both are subtracted (both terms are multiplied by -1), then the numerator must be < 0; $C_4 \ > 0$ and doesn't affect the sign. Overall then, the fraction is < 0. The second term in the equation is just 1 and is being subtracted, and the third term, also a fraction like the first term, is > 0, since $x \ > \ 0$ - and, again, its squared anyway - and $C_3 \ > \ 0$, this positive value is being subtracted. From what I can see, all the terms are "positive terms being subtracted" (I suppose a better (more correct?) way of saying it is all the terms are negative, since there is actually a -1 in each term). From this, I conclude that $f'(x) \ <= \ 0$; therefore, the function, $f(x)$ is monotonically decreasing.

Is this accurate? I like to look at the terms without the sign in front, try to figure out if they are, for example, always positive or negative, and then to take into account the sign in front of the whole term: for example $g(x) = h(x) - y(x)$ if $h(x) <= 0$, then the first term has a positive sign in front, and is therefore negative, and if $y(x) >= 0$, then the second term is "overall negative", since it has a negative sign in front; therefore $g(x) <= 0$. I hope that makes sense. Is that a good method of approaching these things?

Any help much appreciated.

 

[RUber]

Generally, these are all good methods for checking if a function is increasing or decreasing. Another option would be to put in x=0 and x=infinity. The limit as x goes to zero is positive infinity and the limit as x goes to infinity is negative infinity.

 

[pbuk]

That looks like an expression related to a rate of reaction. If a rate of reaction is monotonic, is it necessary to state whether it is increasing or decreasing?

 

[Qwertywerty]

Is this accurate? I like to look at the terms without the sign in front, try to figure out if they are, for example, always positive or negative, and then to take into account the sign in front of the whole term: for example g(x)=h(x)−y(x) g(x) = h(x) - y(x) if h(x)<=0 h(x) = 0 , then the second term is "overall negative", since it has a negative sign in front; therefore g(x)<=0 g(x)

Another option would be to put in x=0 and x=infinity. The limit as x goes to zero is positive infinity and the limit as x goes to infinity is negative infinity.

@nobahar , yes what you reason is fine . One thing that you should note , however , is that in the example you have used , you are supposed to consider only x > 0 , not equal to ( From your third term ) .
You need to always be careful while using the d/dx operation , keeping in mind that the function must be differentiable in the domain being considered .

@RUber , your point is not convincing . A function with value +∞ at x=0 , and -∞ at x→∞ , doesn't necessarily have to be monotonic .
Edit : Unless I have misinterpreted your statement .

Hope this helps ,
Qwertywerty .

 

[nobahar]

Thanks for the response, RUber. I never thought of doing that; I guess if you know that the function is monotonic, then that would be a quick way of identifying whether it is increasing or decreasing. I wanted to identify if it was also monotonic. It is reassuring that both methods agree that it is decreasing.

MrAnchovy, the equation is based on equilibrium constants - which I replaced with C to make the equation "clearer" - and equilibrium concentrations. It is derived from mass and charge balance equations, and used to find the hydrogen ion concentration at equilibrium. I have to find f(0); I am guessing it is important for the equation to be monotonic (at least locally) to facilitate the process of finding f(0) using Newton's method. Although, I don't suppose it matters if it is increasing or decreasing; I just feel more comfortable if I know what the equation is doing a little better: it looks a little daunting, otherwise.

 

[nobahar]

@nobahar , yes what you reason is fine . One thing that you should note , however , is that in the example you have used , you are supposed to consider only x > 0 , not equal to ( From your third term ) .
You need to always be careful while using the d/dx operation , keeping in mind that the function must be differentiable in the domain being considered.

Regarding x=0: Thanks qwerty, I didn't notice that; I need to be more careful.
Regarding differentiability: Thanks for the information; I'll need to go and look into that a bit more as I am not that familiar with it (I vaguely remember something about it...!).