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 [tex]5^x + 7^x=12^x[/tex] x=1 btw- how to solve monotonius functions?
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 [tex]5^x + 7^x=12^x[/tex] 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.
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 [tex]\frac{5^x}{12^x}[/tex], [tex]\frac{7^x}{12^x}[/tex] and some other function like [tex]5^2^x=5^1[/tex]
[itex]f(x)= \frac{5^x}{12^x}= \left(\frac{5}{12}\right)^x[/itex] and [itex]g(x)= \frac{7x}{12^x}= \left(\frac{7}{12}\right)^x[/itex] are functions. It would make no sense to say "solve for x" since x could be any number. The last, [itex]5^{2x}= 5[/itex] is an equation (the "=" between two different functions is a give-away!). It is only true if x= 1/2.
but also [itex]f(x)=5^2^x[/itex] is monotonic function, so also [itex]5^x[/itex](lets say [itex]5^x[/itex] instead of 5) must bee monotonic function. [tex]5^2^x=5^x[/tex]
Make it simpler, take [itex]f(x)=x[/itex] and [itex]f(x)=2x[/itex]. What about that? Should I dare to write [itex]x = 2x[/itex]? 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.
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? )
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).
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.
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?
No, I am not talking about this function, I am talking about some function else, like: [tex]5^x+5^2^x-6=0[/tex]