Given the function f(x) = 1/(x-1) + 1/(x-2) + ...+1/(x-2009), determine the number of real solutions of the equation f(x)=a.
We'll use Rolle's theorem to determine the number of real solutions of the equation.
For this reason, we'll differentiate with respect to x:
f'(x) = -1/(x-1)^2 - 1/(x-2)^2 - ... - 1/(x-2009)^2
We notice that the 1st derivative is negative for any real value of x, therefore f(x) is decreasing over each of these intervals (-infinite,1) ; (1,2) ; (2,3); .....; (2009,+infinite).
If we'll draw the graph of the function we can find the number of roots just counting the intercepting points between the graph of the function and the line x = a.
If "a" belongs to the interval (0 ; +infinite), then the equation has 2009 solutions. If "a" belongs to the interval (-infinite,0), then the equation has 2009 solutions, also.
Therefore, the equation f(x)=a has 2009 solutions.