Why Do Monotonous Functions Have Only One Unique Solution?

[Physicsissuef]

Why there is only one solution of the monotonous functions, when all of the functions all monotonous?

For example, I read that this example is monotonius function and because of that have only one solution

5^x + 7^x=12^x

x=1


btw- how to solve monotonius functions?

 

[HallsofIvy]

In English that would be "monotonic". What do you mean by "all of the functions are monotonic? And my problem is not with your saying that 5^x + 7^x=12^x is monotonic but that it is a function at all! That is an equation. Perhaps you are confusing "function" with "equation" but in that case, I don't recognise the phrase "monotonic equation".

In any case, the definition of "monotonic" is that if f(x)= f(y) then x= y. You cannot have two different values of the independent variable giving the same value of the dependent variable.

That means, in particular, that if f(x) is a monotonic function then the equation f(x)= a, for any a, cannot have more than one solution- it has either no solutions or one.

There is no general method for solving equations involving monotonic functions. Many polynomial functions, of odd degree, are monontonic but there is no general method of solving them.

 

[Physicsissuef]

Yes, you are tottally right. I mistranslated all the things, and you are reading my mind... I was supposed to say, that every function is either monotonically increasing or monotonically decreasing. What is the difference between the functions \frac{5^x}{12^x}, \frac{7^x}{12^x} and some other function like 5^2^x=5^1

 

[HallsofIvy]

Yes, you are tottally right. I mistranslated all the things, and you are reading my mind... I was supposed to say, that every function is either monotonically increasing or monotonically decreasing. What is the difference between the functions \frac{5^x}{12^x}, \frac{7^x}{12^x} and some other function like 5^2^x=5^1

f(x)= \frac{5^x}{12^x}= \left(\frac{5}{12}\right)^x and g(x)= \frac{7x}{12^x}= \left(\frac{7}{12}\right)^x are functions. It would make no sense to say "solve for x" since x could be any number. The last, 5^{2x}= 5 is an equation (the "=" between two different functions is a give-away!). It is only true if x= 1/2.

 

[Physicsissuef]

I can't understand, why monotonic function have only one solution?

 

[Physicsissuef]

but also f(x)=5^2^x is monotonic function, so also 5^x(lets say 5^x instead of 5) must bee monotonic function.

5^2^x=5^x

 

[trambolin]

Make it simpler, take f(x)=x and f(x)=2x. What about that? Should I dare to write x = 2x?

Are you looking for the intersection points of these functions? Because it seems like you are. Then, you can look for the unique arguments of these functions at the intersection points since they are monotonic.

 

[Physicsissuef]

No, I can't understand why monotonic function have only one solution?

 

[HallsofIvy]

If f(x)= f(y), which is larger, x or y?

 

[tiny-tim]

No, I can't understand why monotonic function have only one solution?

Hi Physicsissuef!

"monotonic increasing" means that, if x < y, then f(x) < f(y).

So if f(a) = 0, then f(x) < 0 for all x < a, and f(x) > 0 for all x > a.

In other words, a is the only value of x for which f(x) = 0.

(If you draw a graph, isn't that obvious? )

 

[Physicsissuef]

But isn't all function monotonic increasing, or monotonic decreasing?

 

[tiny-tim]

But isn't all function monotonic increasing, or monotonic decreasing?

Hi !

"monotonic increasing" means that, if x < y, then f(x) < f(y).

"monotonic decreasing" means that, if x < y, then f(x) > f(y).

In other words, the graph of a monotonic increasing function always goes up ,
but the graph of a monotonic decreasing function always goes down .​

"monotonic" doesn't mean single-valued (though the inverse of a monotonic function will be single-valued, over its range).

 

[Physicsissuef]

Hi !

"monotonic increasing" means that, if x < y, then f(x) < f(y).

"monotonic decreasing" means that, if x < y, then f(x) > f(y).

In other words, the graph of a monotonic increasing function always goes up ,
but the graph of a monotonic decreasing function always goes down .​

"monotonic" doesn't mean single-valued (though the inverse of a monotonic function will be single-valued, over its range).

then why everywhere that because of monotonic function it has only one solution?

 

[tiny-tim]

then why everywhere that because of monotonic function it has only one solution?

Because, if y = f(x) is monotonic, it can only cross y = 0 once.

So there is only one value of x for which y = 0.

 

[Physicsissuef]

But it also have one value for x5^2^x-5^x=0

 

[tiny-tim]

But it also have one value for x5^2^x-5^x=0

I don't understand.

Are you saying that 5^2^x - 5^x is monotonic increasing, but that 5^2^x - 5^x = 0 has more than one solution?

 

[Physicsissuef]

I don't understand.

Are you saying that 5^2^x - 5^x is monotonic increasing, but that 5^2^x - 5^x = 0 has more than one solution?

Yes. Some functions are monotonic increasing or decreasing, but they have 2 or more solutions.

 

[tiny-tim]

ok, what are the two solutions of 5^2^x - 5^x = 0?

 

[Physicsissuef]

ok, what are the two solutions of 5^2^x - 5^x = 0?

No, I am not talking about this function, I am talking about some function else, like:

5^x+5^2^x-6=0

 

[tiny-tim]

ok, what are the two solutions of that?

 

[Physicsissuef]

ok, what are the two solutions of that?

y^2+y-6=0

y_1=2

y_2=-3

Ok, I wrong about this one.
But here is another one:

3^x^2^-^\frac{5x}{7}-3^\frac{2}{7}=0

the solutions are:

x_1=1 ; x_2=\frac{-2}{7}

 

[Physicsissuef]

Or this one:
(x-2)^x^2^-^3^x^-^4-1=0

it is x^2 up in there. Also for the previous example it is x^2 instead of x2

 

[tiny-tim]

Hi Physicsissuef!

If you mean 3^(x^2 - (5x/7)) - 3^(2/7), that obviously is not monotonic - it is +∞ for both x = -∞ and +∞.

And the same for (x-2)^(x^2 - 3x - 4) - 1.

(btw, use more brackets, and less ^s)

Look, Physicsissuef, this is a waste of your time - if it's monotonic, it can't be 0 for two different values!

 

[Physicsissuef]

3^(x^2 - (5x/7)) - 3^(2/7), have 2 values which will give us zero. Only two, not more.

 

[tiny-tim]

… "monotonic" means either "always down" or "always up" … !

3^(x^2 - (5x/7)) - 3^(2/7), have 2 values which will give us zero. Only two, not more.

Yes, but it's not monotonic!

It stats at +∞, it goes down, and then it goes up, to +∞.

"monotonic" means either "always down" or "always up"!

 

[Physicsissuef]

Ok, I understand now. Thanks. And is 5^x+7^x-12^x=0, monotonic decreasing or increasing?

 

[tiny-tim]

I've no idea!

Why?

 

[trambolin]

It is zero at negative infinity, nonzero until x=1 and then nonzero again. Thus, it cannot be monotonic. Because if it was monotonic, it should not come back to zero once it is nonzero, which gives you another hint about the derivative of such functions.

But keep in mind, that some people consider constant functions as monotonic, and distinguishing the nonconstant monotonic functions as "strictly" increasing and decreasing

 

[HallsofIvy]

Ok, I understand now. Thanks. And is 5^x+7^x-12^x=0, monotonic decreasing or increasing?

Why did you think 5^x+ 7^x- 12^x is monotonic?

 

[Physicsissuef]

because of my book. It said like that.

5^x+7^x=12^x is monotonic function, that's why it has one solution

 

[Physicsissuef]

I don't know actually how to know which function is monotonic and which is not. Can you help, please?

 

[tiny-tim]

I don't know actually how to know which function is monotonic and which is not. Can you help, please?

The definition is:

"monotonic increasing" means that, if x < y, then f(x) < f(y).

"monotonic decreasing" means that, if x < y, then f(x) > f(y).

In other words, the graph of a monotonic increasing function always goes up ,
but the graph of a monotonic decreasing function always goes down .​

You can generally tell by differentiating: if the derivative is positive everywhere, then the function obviously increases; if it's negative everywhere, it decreases.

 

[Physicsissuef]

So 5^x+7^x=12^x is not monotonic, because if I get for x=1, y=0. So it doesn't go "upward" nor "backward", right?

 

[trambolin]

You are thinking way too complex for this matter. Just reset your mind and try to visualize...

You have a graph. Think about f(x)=x. Plot it with something and ask this "Does this graph has any chance that after some x, the plot bends and starts to decrease?"

Decreasing means "for some x_1 we have f(x_1) and for some x_2 \geq x_1 in other words, for some x_2 sits on the right hand side of x_1, we have f(x_2), that is less than f(x_1)".

Remember, we are always going towards the positive x by convention... So upward, backward etc. forget about all this. If you save your money and you don't spend it, but once in a while you save some more, the amount can only increase, even if you go from negative(paying borrowed money) to positive(saving!), it is still increasing, which means along the way you can cross zero. that is an example for a monotonic function. But if you cross twice, that means you spend some along the way, thus the amount has decreased somewhere! It is also valid if you are always spending but not saving a dime, thus both directions are valid.

Let me emphasize once again. It is about the functions not about the equations. Equations give you the intersection points of the right and left hand sides,

So, here is another statement for you, the function f(x) = 5^x+7^x-12^x is monotonic between [1,\infty)

What do you think?

 

[Physicsissuef]

I would also say that it is monotonic in (-\infty,1], right?

 

[trambolin]

Almost correct. Be careful, something happens at around 0.5, if you can plot it via some software or even with some calculator you will see it.

 

[Physicsissuef]

yes,
x=-2, y=0.0534
x=-1, y=0.259
x=0,5, y=4.88

x=1 , y=0

so it is monotonic increasing for <br /> (-\infty,1)<br />
And how will I know from which point to which one is monotonic increasing, and decreasing? Should I look first for the 0?

 

[trambolin]

First of all, you cannot write (-\infty,1), I understand that you are trying to say somewhere before 1. But what you write is all points from negative infinity to 1 excluding only 1. These are very dangerous mistakes, that you cannot do even when you are asleep!

Regarding your question, just check your notes for finding maxima and minima of functions. I think that is a fair hint for it.

 

[Physicsissuef]

First of all, you cannot write (-\infty,1), I understand that you are trying to say somewhere before 1. But what you write is all points from negative infinity to 1 excluding only 1. These are very dangerous mistakes, that you cannot do even when you are asleep!

Regarding your question, just check your notes for finding maxima and minima of functions. I think that is a fair hint for it.

on this fuction minima is x=1, but maxima I don't know... Why (-\infty,1) (moonotonic decreasing) is not correct?