Prove $ \ \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} \ = \ 0 \ $ has a solution in $ \ (-1,1) $ If $a$ and $b$ are positive numbers, prove that the equation
$$\frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} = 0$$
has at least one solution in the interval $ \ (-1,1) \ $ .
The question is from the exercises section of a textbook chapter on limits/continuity.
I've been stumped on this one for a couple of days. I've been trying to calculate $\lim _{x \to -1}$ and $\lim _{x \to 1}$ and then show the function is continuous to show a root must lie in the interval. Factorising the denominators gives...
$$\frac{a}{(x+1)(x^2+x-1)} + \frac{b}{(x-1)(x^2+x+2)} = 0$$
So of course $x = 1$ and $x = -1$ are undefined and so the limits will be one-sided. Playing around with equation I haven't been able to find an equivalent function across $x \neq -1, x \neq 1$.
The only thing I have been able to show is
$$\frac{a}{b} = - \frac{(x+1)(x^2+x-1)}{(x-1)(x^2+x+2)}$$
and so
$$\lim _{x \to -1} \frac{a}{b} = 0, \lim _{x \to 1} \frac{b}{a} = 0$$
but I'm not sure if this is significant or I'm overthinking things. Could anyone point me in the right direction?
 A: To solve this problem, first of all think about the x values for which the following function is undefined.
We have,
$$f(x) = \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2}$$
which can be written as,
$$f(x) = \frac{a}{(x+1)(x^2+x-1)} + \frac{b}{(x-1)(x^2+x+2)}$$ 
Note that the above function is undefined for
$$x = \pm 1\ , \ \frac{-1\pm \sqrt{5}}{2} \ ...(1)$$
Our interval is $(-1,1)$, so forget about $\frac{-1 - \sqrt{5}}{2}$ which is approximately $-1.62$.
Now, let's calculate the following limits,
$$\lim_{x \to 1^{-}}f(x) =\lim_{x \to 1^{-}}\left( \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2}\right) = -\infty $$ as the first term is simply $\frac{a}{2}$ and the function $x^3 + x - 2$ goes to $0$ from behind as x approaches $1$ from the left. So, the second term in the function is a huge negative number.
And,
$$\lim_{x \to {\left( \frac{-1 + \sqrt{5}}{2}\right)}^{+}}f(x) =\lim_{x \to {\left( \frac{-1 + \sqrt{5}}{2}\right)}^{+}}\left( \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2}\right) = \infty $$ as the function $x^3 + 2x^2 - 1$ approaches $0$ from above from x approaches $\frac{-1 + \sqrt{5}}{2}$ from the right. So, the first term is actually a big positive number and the second term which is actually defined is negligible.
Now, the reason for calculating these limits was to get to the answer by saying that between x = $\frac{-1 + \sqrt{5}}{2}$ and x = $1$, $f(x)$ is continuous everywhere (see $(1)$). The limits for these values of x show that the function has to cross the x-axis between these values of x, that is, there should be at least one root!
The given graph for $ \frac{2}{x^3 + 2x^2 - 1} + \frac{3}{x^3 + x - 2}$
  illustrates the above proof:

A: While the "theme" of this problem is application of the Intermediate Value Theorem, the function involved contains an interesting complication, for which one must be alert.  The quadratic polynomial $ \ x^2 \ + \ x \ + \ 2 \ $ is "irreducible" in the real numbers, but the other quadratic factor appearing here, $ \ x^2 \ + \ x \ - \ 1 \ $ , has the zeroes $ \ \frac{-1 \ \pm \sqrt{5}}{2} \ $ .  Since these are "named numbers" related to the "Golden Ratio" $ \ \phi \ $ , we may use the "shorthand" that the zeroes are $ \ -\phi \ \approx \ -1.618 \ $ and $ \ \frac{1}{\phi} \ \approx \ 0.618 \ $  .  The full factorization of the rational functions is then
$$ \frac{a}{(x+1)(x + \phi)(x - \frac{1}{\phi})} \ + \ \frac{b}{(x-1)(x^2+x+2)} \ = \ 0 \ . $$
The domain of this function on the left-hand side of the equation is then
 $$ \ (-\infty, \ -\phi) \ \cup \ (-\phi, \ -1) \ \cup \ (-1, \ \frac{1}{\phi}) \ \cup \ (\frac{1}{\phi}, \ 1) \ \cup \ (1, \ \infty) \ . $$  As the two terms in the sum are each rational functions, the function in question is continuous everywhere on this domain.  We shall only be concerned in this problem with the function's "behavior" on $ \ (-1, \ \frac{1}{\phi}) \ \cup \ (\frac{1}{\phi}, \ 1) \ $ .
We are given that $ \ a \ $ and $ \ b \ $ are positive.  On $ \ (-1, \ \frac{1}{\phi}) \ $ , the factors $ \ (x - \frac{1}{\phi}) \ $ and $ \ (x - 1) \ $ are negative and the rest are all positive (indeed, $ \ x^2 + x + 2 \ $ is always positive).  Hence, both rational functions are negative on this interval, making the entire function negative there.  So any zero the function may have is not to be found there.
On $ \ (\frac{1}{\phi}, \ 1) \ $ , the rational function of the first term becomes positive, so we now have a sum of terms with opposite signs.  This alone will not prove that our function has an $ \ x-$ intercept.  We may note, though, that the "behavior" of the function in the neighborhood of $ \ x \ = \ \frac{1}{\phi} \ $ is
$$ \frac{a}{(\frac{1}{\phi}+1)(\frac{1}{\phi} + \phi)(x - \frac{1}{\phi})} \ + \ \frac{b}{(\frac{1}{\phi}-1)([\frac{1}{\phi}]^2 + \frac{1}{\phi} + 2)} \ \ \longrightarrow \ \ \frac{K}{  (x - \frac{1}{\phi})} \ + \ L  \ \ , $$
with $ \ K \ > \ 0 \ $ and $ \ L \ $ being constants.  In the neighborhood of $ \ x \ = \ 1 \ $ , the function behaves like
$$ \frac{a}{2} \ + \ \frac{b}{4 \ (x-1) }   \ \ . $$ 
The one-sided limits for the function near the asymptotes are then
$$ \ \lim_{x \ \rightarrow \ \frac{1}{\phi}^{+}} \ f(x) \ = \ +\infty  \ \ \text{and} \ \ \lim_{x \ \rightarrow \ 1^{-}} \ f(x) \ = \ -\infty  \  \ , $$
for which I am using the (not-universal) shorthand notation of "infinite limits".  
Our function is continuous on $ \ (\frac{1}{\phi}, \ 1) \ $ , and tends to "opposite-signed infinities" at either "end" of this interval.  Hence, by the Intermediate Value Theorem, there must be a value of $ \ x \ $ in this interval at which $ \ f(x) \ = \ 0 \ $ .  
[Concerning my earlier comment on the problem  , since $ \ \lim_{x \ \rightarrow \ -1^{+}} \ f(x) \ $ is also $ \ -\infty \ $ , were it not for the asymptote at $ \ \frac{1}{\phi} \ $ , the Intermediate Value Theorem wouldn't be of much help.]
A: In the interval $(-1,1)$ your equation is equivalent to 
$f(x):= a(x-1)(x^2+x+2)+b(x+1)(x^2+x-1)=0$
Now you can see that $f(x)$ is continuous and 
$\lim\limits_{x\to -1}f(x)<0, \quad 
\lim\limits_{x\to 1}f(x)>0$
Thus $f(x)=0$ has at least one solution (in the interval $(-1,1)$).
A: Proof:
Let $f(x)=\frac{a}{x^{3}+2 x^{2}-1}+\frac{b}{x^{3}+x-2} .$ We want to prove that $f$ has at least one solution in the interval
$(-1,1) .$ It suffices to prove that $f(x)=0 .$ By factorization, $f(x)=\frac{a}{(x+1)(x-a)(x-b)}+$$\frac{b}{(x-1)\left(x^{2}+x+2\right)} .$ We denote $\frac{-1+\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2}$ as $a, b$ respectively. We find $x$ such that $f$ is discontinuous. Clearly, $x=\pm 1, a, b .$ However, since the given interval is $(-1,1)$ then $f$ is discontinuous at $x=a .$ Now, suppose $r, s$ is in $(a, 1)$ with $r>a$ and $s<1$ Clearly, $f$ is continuous on $[r, s] .$ Note that, $\lim _{x \rightarrow a^{+}} f(x)=\infty$ and $\lim _{x \rightarrow 1^{-}} f(x)=-\infty .$ Hence, by the Intermediate Value Theorem, since $f$ is continuous on the closed interval $[r, s]$ and previously we have shown that $-\infty<f(x)<\infty$ then there exist a $c \in(r, s)$ such that $f(c)=0$. Hence,
$f(x)$ has at least one solution in $(-1,1)$ namely, $x=c$.
