Continuity of function

Define $f(x) = \frac{1}{x-a_1} + \frac{1}{x-a_2} + \cdots + \frac{1}{x-a_n}.$

• $\displaystyle f(x) \prod_{k=1}^n (x-a_k) = 0$ is a polynomial equation of degree $n-1,$ so we can conclude that $f(x) = 0$ has at most $n-1$ real roots.
• For $1\leq k \leq n-1,$ we can see that $f(x)$ is continuous on $(a_k, a_{k+1})$ and $\lim_{x\to a_k^+} f(x) = +\infty \\ \lim_{x\to a_{k+1}^-} f(x) = -\infty$ so by the intermediate value theorem, there is at least one real root in each $(a_k, a_{k+1}),$ so $f(x) = 0$ has at least $n-1$ real roots.

Together, this shows $f(x) = 0$ has exactly $n-1$ real roots (and further, every root of $f(x)=0$ is simple)

Let f(x)= $\frac{1}{x-a_1}$ + $\frac{1}{x-a_2}$ +...+ $\frac{1}{x-a_n}$ , than there exists n-1th polynomial g(x) such that g(x)=( $x-a_1$ )( $x-a_2$ )...( $x-a_n$ )f(x). Than g( $a_n$ )>0, g( $a_{n-1}$ )<0, g( $a_{n-2}$ )>0 ... Therefore, according to the intermediate value theorem, g(x)=0 must have at least one real root between each interval ( $a_1$ , $a_2$ ), ( $a_2$ , $a_3$ ),..., ( $a_{n-1},$ $a_n$ ). But since the number of real roots of g(x) cannot be greater than n-1, the number of the real roots of g(x) is n-1. Also, since the roots of g(x) do not make f(x)'s denominator 0, the number of the real roots of f(x) is n-1.

If a polynomial has degree n, then it will have n real solutions(if all it's coefficients are real).

In this polynomial, the function has degree n-1, so it will have n-1 solutions.