(a) Prove the the equation $$x\cdot\left(1+\sqrt{x^2+1}\right)^3=\frac{1}{2}$$ has a unique solution in $\mathbb{R}$.
(b) Let $a_1,a_2,a_3,b_1,b_2,b_3\in\mathbb{R}$ such that $a_1,a_2,a_3>0$ 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 $\mathbb{R}$.
Can I have feedback on my proofs to see that I'm going in the right directions? Also in the second one I'm a bit stuck..
Let $f:\ R➜R,\ f\left(x\right)\ =\ x\left(1+\sqrt{x^{2}+1}\right)^{3}$.
Assume toward contradiction there exists $a,b$ such that $f(a)=f(b)=\frac{1}{2}$
Note that by the algebra of continuous functions f is continuous on [a,b].
Also by the algebra of differentiable functions f is differentiable on (a,b).
Therefore we can compute $f'\left(x\right)$
$f'\left(x\right)=1\cdot\left(1+\sqrt{x^{2}+1}\right)^{3}+3\left(\frac{2x}{2\sqrt{x^{2}+1}}\right)\cdot x=\left(1+\sqrt{x^{2}+1}\right)^{3}+3\left(\frac{2x^{2}}{2\sqrt{x^{2}+1}}\right)$
Note that $f'\left(x\right) > 0$ for every $x ∈ R$.
But by Rolle's theorem there exists a point $a < c < b$ such that $f'\left(c\right) = 0$, which means we have a contradiction!
Therefore there exists a unique solutions to $f(x)=\frac{1}{2}$.
Note that
$\frac{a_{1}}{x-b_{1}}+\frac{a_{2}}{x-b_{2}}+\frac{a_{3}}{x-b_{3}}=0 \ \ \ $ ➜$ \ \ \ a_{1}\left(x-b_{2}\right)\left(x-b_{3}\right)+a_{2}\left(x-b_{1}\right)\left(x-b_{3}\right)+a_{3}\left(x-b_{1}\right)\left(x-b_{2}\right)=0$
Let Let $f:\ R➜R,\ f\left(x\right)\ = \ a_{1}\left(x-b_{2}\right)\left(x-b_{3}\right)+a_{2}\left(x-b_{1}\right)\left(x-b_{3}\right)+a_{3}\left(x-b_{1}\right)\left(x-b_{2}\right)$
Note that $b_{1}<b_{2}<b_{3}$ and $a_{1},a_{2},a_{3}\ >\ 0$, and that by the algebra of continuous functions $f$ is continuous. Therefore we have
$f\left(b_{1}\right)\ =\ a_{1}\left(b_{1}-b_{2}\right)\left(b_{1}-b_{3}\right)+a_{2}\left(b_{1}-b_{1}\right)\left(b_{1}-b_{3}\right)+a_{3}\left(b_{1}-b_{1}\right)\left(b_{1}-b_{2}\right)=a_{1}\left(b_{1}-b_{2}\right)\left(b_{1}-b_{3}\right)+0+0\ >\ 0$
$f\left(b_{2}\right)\ =\ a_{1}\left(b_{2}-b_{2}\right)\left(b_{2}-b_{3}\right)+a_{2}\left(b_{2}-b_{1}\right)\left(b_{2}-b_{3}\right)+a_{3}\left(b_{2}-b_{1}\right)\left(b_{2}-b_{2}\right)=0+a_{2}\left(b_{2}-b_{1}\right)\left(b_{2}-b_{3}\right)+0\ <\ 0$
$f\left(b_{3}\right)\ =\ a_{1}\left(b_{3}-b_{2}\right)\left(b_{3}-b_{3}\right)+a_{2}\left(b_{3}-b_{1}\right)\left(b_{3}-b_{3}\right)+a_{3}\left(b_{3}-b_{1}\right)\left(b_{3}-b_{2}\right)=0+0+a_{3}\left(b_{3}-b_{1}\right)\left(b_{3}-b_{2}\right)\ >\ 0$
Note that Since $f\left(b_{1}\right)\cdot f\left(b_{2}\right)<0$ and $f\left(b_{2}\right)\cdot f\left(b_{3}\right)<0$, by the general intermediate value theorem there exist a point $b_{1}\ <\ c_{1}\ <\ b_{2}$ and $b_{2}\ <\ c_{2}\ <\ b_{3}$ such that $f\left(c_{1}\right)=f\left(c_{2}\right)=0$ and $c_{1}\ \ne\ c_{2}$.
From here I'm a bit stuck on how to prove that the points are unique.. I thought about saying that the function is decreasing/increasing on the intervals but I'm not sure it's true or how to show it