2
$\begingroup$

If a and b are positive numbers, prove that the equation

$$ \frac{a}{x^3+2x-1} + \frac{b}{x^3+x-2}=0$$

has at least one solution in the interval $(-1,1)$.

You can't plug and chug because you'll wind up with an undefined term for both numbers terminating the given interval. We do know that if c exists, at $f(c)$, $x^3 + 2x - 1$ and $x^3 + x - 2$ evaluate to numbers of opposite sign AND that the two terms are equal; otherwise two numbers with positive numerators a and b would not add to zero. And if we assume that c does exist, then $f(-1) < f(0) < f(1)$.

If this is the case, then, for $f(-1)$, the term which evaluates to a greater value must be negative. Similarly, for $f(1)$, either both terms are positive or the term which evaluates to a greater value must be positive.

I'm sorta groping in the dark here, collecting information but unsure of how to apply it to prove the original assertion.

$\endgroup$
4
$\begingroup$

Your equation is equivalent to: $$f(x)=\frac{x^3+2x-1}{x^3+x-2}=-\frac{a}{b}.$$ Notice that $f(x)$ is a continuous function on $[-1,1)$, and since $f(-1)=1$ while $$\lim_{x\to 1^-}f(x)=-\infty,$$ $f(x)$ takes any value in $[0,-\infty)$ over $I=[-1,1)$.

Additionally, $f(x)$ is decreasing over $I$, hence the solution is unique.

$\endgroup$
12
  • 1
    $\begingroup$ You're brilliant dude. Assuming the first term was greater than zero, and thus the second less than zero, I set up an equation using this information, then used the cubic formula to solve for c. Your solution is far more elegant. I hope to attain your level of mathematical maturity at some point. Thank you. $\endgroup$ – akuryo Sep 13 '14 at 6:46
  • $\begingroup$ Actually, wait a second. The exponent attached to the highest terms in both the numerator and denominator are both 3. Beautiful solution, wrong problem. $\endgroup$ – akuryo Sep 13 '14 at 7:13
  • 1
    $\begingroup$ I'm sorry; I don't understand f(1-) = - infinity. Will you explain? How do you substitute 1 from the left? $\endgroup$ – akuryo Sep 13 '14 at 7:35
  • 1
    $\begingroup$ Oh I see. Please excuse all of these questions; I just have a few more. Did you know intuitively that f(1-) = - infinity, or did you make a table of values? If you did know intuitively, will you share your intuition? $\endgroup$ – akuryo Sep 13 '14 at 8:02
  • 1
    $\begingroup$ @akuryo: $1$ is a zero of the denominator, that is negative in a left neighbourhood of $1$. The numerator is positive in a whole neighbourhood of $1$. This gives that the limit of the ratio when $x$ approaches $1^-$ is $-\infty$. $\endgroup$ – Jack D'Aurizio Sep 13 '14 at 8:06
0
$\begingroup$

Hints

  1. $a,b$ are both positive.

  2. Each of the cubics has a real root in the interval $[0,1]$.

  3. In fact it is easy to see that one must be less than the other.

$\endgroup$
2
  • $\begingroup$ I don't understand this at all. $\endgroup$ – TonyK Sep 13 '14 at 7:40
  • $\begingroup$ @TonyK To go a bit further ... One cubic has a root near 0.5, the other has a root at 1. So when you take the reciprocals they go to $\infty$ there. Now look at signs, specifically between the two roots. $\endgroup$ – almagest Sep 13 '14 at 7:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.