Calculus - Rolle question

Question

Regarding:

Let $a_{1},a_{2},a_{3},b_{1},b_{2},b_{3} \in \mathbb{R}$, while $a_{1},a_{2},a_{3}>0$ and $b_{1}<b_{2}<b_{3}$.

Prove that the equation $$\frac{a_{1}}{x-b_{1}}+\frac{a_{2}}{x-b_{2}}+\frac{a_{3}}{x-b_{3}}=0$$

has exactly two distinct solutions in $\mathbb{R}$.

--

So I have tried defining a function $f$ which is equal to the left side of the equation, derivate it, and assume by contradiction that it has more than 2 solution.

Unfortunately with no luck...

If someone knows a way, please only give me a hint so I can solve it by myself.

Thanks a lot!

Answer 1

More generally, suppose there are $n$ $a_k$ and $b_k$ with $a_k>0$ and $0 < b_k < b_{k+1}$ for all $k$.

Let $f(x) =\sum_{k=1}^n a_k/(x-b_k) $.

$f'(x) =-\sum_{k=1}^n a_k/(x-b_k)^2 \lt 0 $ so $f(x)$ is always decreasing wherever it is defined.

If $x=b_k+c$ where $c$ is small, $f(x)>0$ because all the other terms are bounded. Similarly, if $x=b_k-c$ where $c$ is small, $f(x)<0$.

This can be made rigorous by choosing $c_0= \min(b_{k+1}-b_k)/2$, getting an upper bound on each $a_j/(x-b_j)$ for $|x-b_j|<c_0$ with $j\ne k$, and making $x-b_k$ small enough.

Therefore there is exactly one root between each pair of $b_k$.

If $x$ is greater than all the $b_k$, then all the terms are positive so there is no root there.

If $x$ is smaller than all the $b_k$, then all the terms are negative so there is no root there.

Therefore the only roots are between the $b_k$ so there are $n-1$ of them.

Answer 2

We can simplify this equation as: $f(x)=a_1(x-b_2)(x-b_3)+a_2(x-b_1)(x-b_3)+a_3(x-b_1)(x-b_2)$ $f(b_1)=a_1(b_1-b_2)(b_1-b_3)>0,$ $f(b_2)=a_2(b_2-b_1)(b_2-b_3)<0$ $f(b_3)=a_3(b_3-b_1)(b_3-b_2)>0.$ So by IVT, there is one root in $(b_1,b_2)$ and other one is in $(b_2,b_3)$. So only two real root of this equation which is a quadratic.