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.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.

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