0
$\begingroup$

Let $a_1,a_2,a_3,b_1,b_2,b_3\in \mathbb{R}$, $a_1,a_2,a_3>0$, $b_1<b_2<b_3$.
Prove that $\frac{a_1}{x-b_1}+\frac{a_2}{x-b_2}+\frac{a_3}{x-b_3} = 0$ has exactly two solutions in $\mathbb{R}$.

First, I showed that the solutions must be bounded in $[b_1,b_3]$.
Since there are 3 points that the function is undefined at, I couldn't use any theorem that requires the function to be continuous. I know that it's continuous between $(b_1,b_2)$ and $(b_2,b_3)$, so I was trying to use IVT, but it also didn't work.

Could you please give me a hint?

$\endgroup$
4
  • $\begingroup$ You can use IVT on on [b_1+r,b_2-r] for small r and the property that the function is locally monotonous. $\endgroup$
    – TomTom314
    Commented Jan 16, 2021 at 11:51
  • $\begingroup$ @TomTom314 Can I just say "Let $r\in \mathbb R$ be a very small positive number? $\endgroup$
    – Lilo
    Commented Jan 16, 2021 at 12:01
  • $\begingroup$ Yes, I should do that. $\endgroup$
    – Z Ahmed
    Commented Jan 16, 2021 at 12:04
  • $\begingroup$ Here is the SAME question asked and answered yesterday math.stackexchange.com/questions/3986096/… $\endgroup$
    – Raffaele
    Commented Jan 16, 2021 at 13:05

2 Answers 2

1
$\begingroup$

You need to use IVT in the two intervals $(b_1,b_2)$ and $(b_2, b_3)$ near the endpoints.

At $b_1^+$, $\frac{{a_1}}{x-b_1}+\frac{{a_2}}{x-b_2}+\frac{{a_3}}{x-b_3} \to \infty$.

That is because $\frac{{a_1}}{x-b_1}+\frac{{a_2}}{x-b_2}+\frac{{a_3}}{x-b_3}> \frac{{a_1}}{x-b_1}+\frac{{a_2}}{b-b_2}+\frac{{a_3}}{b-b_3} \to \infty$, where $b = \frac {b_1+b_2}2$.

Similarly at $b_2^-$, $\frac{{a_1}}{x-b_1}+\frac{{a_2}}{x-b_2}+\frac{{a_3}}{x-b_3} \to -\infty$.

By IVT, this shows that $\frac{{a_1}}{x-b_1}+\frac{{a_2}}{x-b_2}+\frac{{a_3}}{x-b_3} =0$ for some $x \in (b_1,b_2)$. The result for $(b_2, b_3)$ is similar.

$\endgroup$
0
$\begingroup$

Obviously $x\ne b_1,b_2mb_3$ a simplification gives: $$f(x)=a_1(x-b_2)(x-b_3)+a_2(x-b_1)(x-b_3)+a_3(x-b_1)(x-b_2)$$ So we get $$f(b_1)=a_3(b_1-b_2)(b_1-b_3)>0$$ $$f(b_2)=a_2(b_2-b_1)(b_1-b_3)<0$$ $$f(b_3)=a_3(b_3-b_1)(b_2-b_2)>0$$ By IVT $f(x)=0$ has two real roots one in $(b_1,b_2)$ and one in $(b_2,b_3)$. Effectively, the given eq. is a quadratic which has both rotts real.

$\endgroup$
3
  • $\begingroup$ Aren't $f(b_1), f(b_2), f(b_3)$ undefined since $x\neq b_1,b_2,b_3$? $\endgroup$
    – Lilo
    Commented Jan 16, 2021 at 11:59
  • $\begingroup$ Please mote that my $f(x)$ is quadratic not the original one you may call it $g(x)$ since $b_1,b_2,b_3$ cannot be the roots of this equation. Hence roots of $f(x)=0$ and $g(x)=0$ are the same. $\endgroup$
    – Z Ahmed
    Commented Jan 16, 2021 at 12:02
  • $\begingroup$ Do I need to explain the relation between $g(x)$ and $f(x)$ or is it trivial enough? $\endgroup$
    – Lilo
    Commented Jan 16, 2021 at 12:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .