<s>[INST] Prove that the equation has solutions in two different intervals [-1,1] and [1,2] $a,b,c \in \mathbb{R}$ and $a,b,c$ positive numbers
$\frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=1$
Show that the equation has a solution in the interval $[-1,1]$ and a solution in the interval $[1,2]$.

$\frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=1$
$\iff [(a+b)\cdot x+a-b] \cdot (x-2) + c \cdot (x^2-1) = (x-2)\cdot (x^2-1)$
$\iff [(a+b)\cdot x+a-b] \cdot (x-2) + c \cdot (x^2-1) - (x-2)\cdot (x^2-1) = 0$
$f(x) := [(a+b)\cdot x+a-b] \cdot (x-2) + c \cdot (x^2-1) - (x-2)\cdot (x^2-1)$
$f(-1) = [-a+b+a-b]\cdot (-3)+c \cdot 0-(-3) \cdot 0 = 0$
$f(1) = [a-b+a-b] \cdot (-1)+0-(-1)\cdot 0 = 2b-2a \lt 0$, for $a \gt b$
$f(2) = [(a-b)\cdot 2 +a-b] \cdot 0 +c \cdot 3 -0\cdot 3= 3c \gt0$
How do I go on from here?
Or is this even the right way?
 [/INST] Hint: $$\lim_{x \to -1+} \frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=\infty$$
$$\lim_{x \to 1-} \frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=-\infty$$
$$\lim_{x \to 1+} \frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=\infty$$
and
$$\lim_{x \to 2-} \frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=-\infty$$.
By continuity, there are points in $(-1,1)$ and $(1,2)$ where any real value is attained, in particular the value $1$.
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