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be the set of positive real numbers. Determine all functions 
such that for all positive real numbers 
and 
![\[f(x+f(xy))+y=f(x)f(y)+1\]](http://latex.artofproblemsolving.com/c/6/a/c6a968a17f6fd5fd637c1ee5d855bdff11e5423a.png)
and 
such that![\[
1^n + 2^n + 3^n + \cdots + n^n = x!
\]](http://latex.artofproblemsolving.com/6/4/b/64b5f5919fd3710cc15e96724918eae216c8e4f0.png)
be 
. The tangents to at 
meet at point 
. For a point 
on line 
which is not on the segment , let the midpoint of 
be 
. Lines 
meet again at points 
respectively. Let 
be the midpoint of . Prove that the points 
lie on a circle.
red cards, 
white cards, and 
blue cards. A player plays all the cards one at a time. With each play he accumulates a penalty. If he plays a blue card, then he is charged a penalty which is the number of white cards still in his hand. If he plays a white card, then he is charged a penalty which is twice the number of red cards still in his hand. If he plays a red card, then he is charged a penalty which is three times the number of blue cards still in his hand. Find, as a function of 
and 
the minimal total penalty a player can amass and all the ways in which this minimum can be achieved.
and the 
s are distinct?
, of class 
on its maximal domain 
. We have 
on 
. Compute 
's limits at 
, and its lateral limits at 
, 
. That is enough to prove the thesis.
, as defined, is not a polynomial, but a rational function 
, where 
. And now, if you apply Descartes' rule of signs to 
, even if you get 
changes of signs, that does not guarantee the existence of real roots. Descartes' rule of signs is not that precise!
, as defined, is not a polynomial, but a rational function , where . And now, if you apply Descartes' rule of signs to , even if you get changes of signs, that does not guarantee the existence of real roots. Descartes' rule of signs is not that precise!
.Plugging x=
,we observe that changes its sign 
times. is continuous, so by 
has real roots each in
.Now another root will surely lie in b/w 
& 
. Now for no root the polynomial 
becomes 
.So the given equation ,i.e., the rational function 
has exactly real roots.
& .. & ?
and 
. Then how can u proceed
we clearly see that the given equation can be written as a polynomial in of degree n. Let the polynomial be denoted as 
. It can also be easily seen that none of the 
for every 
. complex roots.(counted with multiplicities).
is a solution
Which after solving reduces to,
This immediately implies that 
is zero and thus the proof that every root is real is complete.
Which is obviously 
. Hence, we can conclude that 
is an increasing function... observe that also has exactly points of discontinuity 
namely at 
.. must have distinct roots each lying between 
.. has 'exactly' these roots and no more? Well, that's trivial.. just consider the fundamental theorem of algebra and you are done 
fall in the domain of 
. It's already clearly implied 
Which is obviously . Hence, we can conclude that is an increasing function... observe that also has exactly points of discontinuity namely at .. must have distinct roots each lying between .. has 'exactly' these roots and no more? Well, that's trivial.. just consider the fundamental theorem of algebra and you are done
.Plugging x=,we observe that changes its sign times. is continuous, so by has real roots each in.Now another root will surely lie in b/w & . Now for no root the polynomial becomes .So the given equation ,i.e., the rational function has exactly real roots.
Which is obviously . Hence, we can conclude that is an increasing function... observe that also has exactly points of discontinuity namely at .. must have distinct roots each lying between .. has 'exactly' these roots and no more? Well, that's trivial.. just consider the fundamental theorem of algebra and you are done 
for 
, for otherwise one or more terms of the left hand side of the given equation is/are undefined...Then we can write the equation as ![$$\sum_{i=1}^{n}a_i[(a_1-x) \cdots \widehat{(a_i-x)} \cdots (a_n-x)]=2015 \prod_{j=1}^{n}(a_j-x)$$](http://latex.artofproblemsolving.com/d/c/9/dc9d139147dc62103622174a663351cd3171e1aa.png)
,where the 
denotes a deleted term..We can also write this equation in form 
where 
and 
are polynomials in ![$\mathbb{R}[x]$](http://latex.artofproblemsolving.com/a/8/8/a88b9f4858016b1771635c611d44b56161a08009.png)
assuming 
,, such that 
and 
.
..(we've 
) roots...
then we have 
and that gives 
a contradiction , which gives the equation has all real roots. degree polynomial which will have roots , as we proved all of them are real we are done. 
then we have and that gives a contradiction , which gives the equation has all real roots. degree polynomial which will have roots , as we proved all of them are real we are done. 
?
,
means the function is infinitely differentiable, some also consider them bounded, but that differs by professor and notation at times (I don't know the general consensus on that).
hence they only consider 
and not the extended real numbers. On this notion I could be wrong, however.
