Question 1061353
<pre>
This problem might as well be stated:

"If the capital of Georgia is Atlanta, then prove the equation 
xlogx = 3-x is satisfied by at least one value of x lying 
between 1 and 3."

That's because "f(x) = (x-3)logx" has no more to do with whether

"the equation xlogx = 3-x is satisfied by at least one value of x 
lying between 1 and 3" than the capital of Georgia being Atlanta 
has to do with it!

Thus we ignore the f(x) part and take the problem to be:

Prove that the equation 

{{{x*log((x)) = 3-x}}} 

is satisfied by at least one value of x lying between 1 and 3.

Consider the function which is continuous on [1,3]

{{{g(x)=x*log((x))-3+x}}}

{{{g(1)=1*log((1))-3+1=-2}}}

{{{g(3)=3*log((3))-3+3=1.4313638...}}}

g(1) is negative and g(3) is positive.

Thus there is some number, say h, on [1,3] such that

{{{g(h)=0}}}

Thus

{{{g(h)=h*log((h))-3+h=0}}}

and

{{{h*log((h)) = 3-h}}}

[Note: If you require that h be an algebraic number then
choose h to be {{{root(5,100)}}}]

Edwin</pre>