Question 1048554
 {{{f(1-x)+(1-x)f(x)=5}}}

====> Replacing x by 1-x, we get  {{{f(x)+xf(1 - x)=5}}}

Using Cramer's rule to find f(x), 


{{{f(x) =( abs((matrix(2,2,5,1,5,x))))/(abs((matrix(2,2,1-x,1,1,x)))) = (5x-5)/(x(1-x)-1) = (5x-5)/(-x^2+x-1)}}}.

(As a check, {{{f(1-x) = (-5x)/(-x^2+x-1)}}}, which is the expression also obtained when x is replaced by 1-x in the preceding expression for f(x).)


===> f'(x) = {{{(5x^2-10x)/(-x^2 + x-1)^2}}}


Setting this derivative equal to 0, we get the critical values x = 0, 2.
By using the first derivative test, we find a local max at x = 0, and a local minimum at x = 2.


At x = 0, f(0) = 5.
At x = 2, f(2) = -5/3.


Therefore, the absolute max value of f(x) is 5, and the absolute min value of f(x) is -5/3.

We confirm by graphing f(x):
{{{graph( 300, 200, -5, 5, -5, 5, (5x-5)/(-x^2+x-1) )}}}