- #1

nobahar

- 495

- 2

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:

[tex]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}}[/tex]

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:

[tex]f'(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}}[/tex]

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.

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:

[tex]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}}[/tex]

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:

[tex]f'(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}}[/tex]

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.

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