0
$\begingroup$

So let's say I have some random equation $6yx^3 - 3yx + 5 = 0$, but could be anything.

How would I go about finding a value for $y$, that makes it so that this equation only holds true if $x$ is some value within the range $[a, b\textbf{]}$.

Also, is it possible to solve such a problem using wolfram alpha? If so, how would I input it.

$\endgroup$
  • $\begingroup$ Why is this tagged multivariable calculus, geometry, and algebraic geometry? Where did you encounter this problem? $\endgroup$ – apnorton Feb 20 '13 at 21:12
1
$\begingroup$

We have the expression $f(x,y)=0$. There is no value $y=c$ such that $f(x, c) = 0$ for all $x \in [a, b]$.

Why? We would be looking for a $c$ such that the polynomial $f(x,c)$ (I'm assuming this is a polynomial based on the question tags) has infinitely many solutions, namely those in $[a, b]$. This violates the fundamental theorem of algebra.

EDIT: As noted in the comments, this isn't valid for the null polynomial. Namely, if your expression is $f(x, y) = x^2y + xy$, $f(x, y)$ has infinitely many solutions for the choice $y=0$.

$\endgroup$
  • $\begingroup$ Ok. I guess I should limit my statement to non-trivial expressions... $\endgroup$ – apnorton Feb 20 '13 at 21:28
  • $\begingroup$ Would it be possible to find a range of c values for y such that f(x, c) = 0 for all a <= x <= b, instead of just for a single value of c? $\endgroup$ – user1855952 Feb 20 '13 at 21:34
  • $\begingroup$ @user1855952 So... for all $c \in [\alpha, \beta]$, there exists a $d \in[a, b]$ such that $f(d, c) = 0$? This is what you're asking? $\endgroup$ – apnorton Feb 20 '13 at 21:40
  • $\begingroup$ yup, is this a solvable problem? $\endgroup$ – user1855952 Feb 20 '13 at 21:43
  • $\begingroup$ @user1855952 Not necessarily. How about the paraboloid $$f(x, y) = x^2 + y^2 + 5$$. No matter how hard you try, you will find no interval $[\alpha, \beta]$ that satisfies the above, as it never intersects the xy plane. $\endgroup$ – apnorton Feb 20 '13 at 21:49

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.