f(x) = ?

[George Saliaris]

Let f : (0, +infinity) -> R differentiable such that e^(f(x)) + f(x) = lnx + x (1)

f(x)= ?

*It's not a homework problem, I am doing it for my own practice and it seems that I have been struggling for quite some hours*

My attempts

1) Differentiate both sides
2) Some algebra stuff like multiplying with the same factor an altered version of (1)
3) e^f(x) = u ,f(x)= u working with inverse funtions
4) Working with initial functions



Could somebody provide me with an insight?

 

[Deleted member 4993]

Let f : (0, +infinity) -> R differentiable such that e^(f(x)) + f(x) = lnx + x (1)

f(x)= ?

*It's not a homework problem, I am doing it for my own practice and it seems that I have been struggling for quite some hours*

My attempts

1) Differentiate both sides
2) Some algebra stuff like multiplying with the same factor an altered version of (1)
3) e^f(x) = u ,f(x)= u working with inverse funtions
4) Working with initial functions
Could somebody provide me with an insight?

What did you get by differentiating both sides?

 

[Steven G]

What does x(1) mean? Does it mean x*1 = x or something else. If it is x I just find it strange to write it as x(1).

 

[George Saliaris]

I denoted e ^(f(x)) + f(x) = lnx + x as relationship (1) or just (1)... Anyway I tried another approach..

let f(x) < lnx for every x Ε R or for an interval let's say C. Using this and that e^(x) is an increasing function I proved this is impossible. I did the same thing if f(x)> lnx... So f(x) = lnx... Is this way correct?

 

[Romsek]

One thing I'd immediately try is to see if you can match up terms on each side. It's certainly not guaranteed to work but it's simple enough that it's worth a quick try.

In this case

[MATH]e^{f(x)}[/MATH] seems unlikely to match with [MATH]\ln(x)[/MATH] so let's suppose that

[MATH]e^{f(x)} = x[/MATH]
taking logs on both sides we get

[MATH]f(x) = \ln(x)[/MATH]
which fortuitously is the remaining term.

and thus [MATH]f(x) = \ln(x)[/MATH] is what we're after.

Again this isn't a rigorous procedure that guarantees and answer but it's worth a quick check.

 

[Steven G]

I can't emphasize how important it is to try what Romsek did. This is a great skill to have and was the exact way I solved this problem. It came together so nicely.

 

[Dr.Peterson]

let f(x) < lnx for every x Ε R or for an interval let's say C. Using this and that e^(x) is an increasing function I proved this is impossible. I did the same thing if f(x)> lnx... So f(x) = lnx... Is this way correct?

It looks like you must have done something like Romsek's "hopeful guess" method, to suppose that ln(x) might be the answer, and are going beyond that to show that this is the only solution. That was lacking in what Romsek said.

It may be helpful if you show us the details of your work, so we can check more easily whether it is correct.

*It's not a homework problem, I am doing it for my own practice and it seems that I have been struggling for quite some hours*

May I ask where the problem came from, and what you want to practice? That is, if it's from a textbook, what topic is it associated with? In any case, I'm curious whether the context would suggest your method or some other.

 

[George Saliaris]

It looks like you must have done something like Romsek's "hopeful guess" method, to suppose that ln(x) might be the answer, and are going beyond that to show that this is the only solution. That was lacking in what Romsek said.

It may be helpful if you show us the details of your work, so we can check more easily whether it is correct.


May I ask where the problem came from, and what you want to practice? That is, if it's from a textbook, what topic is it associated with? In any case, I'm curious whether the context would suggest your method or some other.

1) e^(f(x)) + f(x) = x+ lnx
e^(f(x)) - e^lnx = lnx - f(x) (2)


Suppose f(x) > lnx for every x>0 then e^(f(x)) > e^lnx (3)

(e^x is an increasing function)

we also have lnx-f(x) < 0 (4)


(2),(3),(4) contradiction


Same way when f(x) < lnx for every x>0
f(x) > lnx or f(x) < lnx at some interval C
f(x) > lnx or f(x) < lnx for ''some x1 ,x2,......xn (not at an interval but at some ''points''


2) It came from my teacher's pdf file on calculus..The exercise belongs to a series of other exercises called ''general exercises on derivatives''..

 

[Dr.Peterson]

1) e^(f(x)) + f(x) = x+ lnx
e^(f(x)) - e^lnx = lnx - f(x) (2)


Suppose f(x) > lnx for every x>0 then e^(f(x)) > e^lnx (3)

(e^x is an increasing function)

we also have lnx-f(x) < 0 (4)


(2),(3),(4) contradiction


Same way when f(x) < lnx for every x>0
f(x) > lnx or f(x) < lnx at some interval C
f(x) > lnx or f(x) < lnx for ''some x1 ,x2,......xn (not at an interval but at some ''points''


2) It came from my teacher's pdf file on calculus..The exercise belongs to a series of other exercises called ''general exercises on derivatives''..

There's some good thinking there. Pondering it, though, I think it can be simplified a lot. You can just apply the same thinking to any particular value of x (rather than an interval) and determine that f(x) = ln(x). That makes me wonder whether the condition of differentiability is needed, or even continuity. The only calculus I see as being needed is that e^x is an increasing function.

 

[George Saliaris]

This 'exercise' is part of another ''larger exercise'' which involves 1) extremums 2) curvature 3) some inequalities to show like (b/a)^(c-b) > (c/b) ^ (b-a) 0<a<b<c ...etc

 

[Romsek]

as you all are still going on about this how about this since you're playing with derivatives.

[MATH] e^f + f = \ln(x) + 1\\ \text{differentiate both sides}\\ f^\prime e^f + f^\prime = \dfrac 1 x + 1\\ f^\prime(e^f + 1) = \dfrac 1 x (x+1)\\ \text{Again we match up the forms and it's immediately suggested that}\\ f^\prime = \dfrac 1 x\\ f = \ln(x) [/MATH]
But I suppose this is just more hopeful guessing.

 

[Dr.Peterson]

But I suppose this is just more hopeful guessing.

The usual term for "hopeful guessing" as I used it here is "by inspection". It's an excellent method to obtain a solution; all that's needed to finish is a proof of uniqueness (or in other cases, to find any other solutions there may be), which can often, as here, be found by using the first solution.