proof using intermediate theorem

[njama]

1. The problem statement, all variables and given/known data

Prove that if c and d are positive, then the equation

\frac{c}{x-1} + \frac{d}{x-3}=0

has at least one solution in the interval (1,3)

2. Relevant equations
Intermediate Theorem

If f is continuous on [a,b], and if f(a) and f(b) are nonzero and have opposite signs, then there is at least one solution of the equation f(x)=0 in the interval (a,b).


3. The attempt at a solution

The problem is that I got open interval (i.e not closed), (1,3).

I started by proving that the function above is continuous on (1,3).
\lim_{x\rightarrow k}\frac{c}{x-1} + \frac{d}{x-3}=\lim_{x\rightarrow k} \frac{c(x-3)+d(x-1)}{(x-1)(x-3)}= \lim_{x\rightarrow k}\frac{x(c+d)-3c-d}{(x-1)(x-3)}

Now because there is open interval, k \neq 1,3 and

\lim_{x\rightarrow k}\frac{x(c+d)-3c-d}{(x-1)(x-3)} = \frac{k(c+d)-3c-d}{(k-1)(k-3)}

Hence I proved that the function \frac{c}{x-1} + \frac{d}{x-3} is continuous on the interval (1,3).

Now the problem is to prove that there is at least one solution in the interval (1,3). If it was closed interval, it would be much easier, since the only condition would need to be satisfied would be to check f(1)>0 and f(3)<0 or vice verse. But the values 1,3 are not in the domain of the function.

Please help me!

[Mark44]

I think you can do this problem without either limits or the Intermediate Value Theorem, which doesn't apply anyway, since you don't have the closed interval [1, 3].

Start with
\frac{c}{x-1} + \frac{d}{x-3}~=~0
Both fractions are defined for all x in (1, 3).

Combine the fractions into a single fraction:
\frac{c(x - 3) + d(x - 1)}{(x - 3)(x-1)}~=~0
Now solve for x.

[njama]

Thanks. I know that I can solve the equation that way, so that

x=\frac{3c+d}{c+d}

Now because c>0 and d>0, c+d>0, so that c+d \neq 0, and there is only one solution of the equation. But this task is under the section of Intermediate theorem, so I am supposed to solve it using the theorem.

My textbook is Calculus (Seventh Edition) by Howard Anton, Irl Bivens and Stephen Davis.

[Mark44]

Just because the problem is in the Intermediate Value Theorem section does not necessarily mean you have to use that theorem. As I said before, the conditions of that theorem aren't met. Your interval is (1, 3), not [1, 3], and you can't evaluate f(1) and f(3) to see if they are of different signs.

If the problem doesn't explicitly say you have to use a particular technique, then don't assume you have to. Anyway, I provided a way to reach the desired conclusion. Take it or leave it.

[emyt]

your continuity "proof" isn't even a proper proof

anyway, Mark is probably right, I've seen in some textbooks, they have "irrelevant" questions because they want to remind you of something that you may have to use for a more relevant question in the same section

[njama]

your continuity "proof" isn't even a proper proof

anyway, Mark is probably right, I've seen in some textbooks, they have "irrelevant" questions because they want to remind you of something that you may have to use for a more relevant question in the same section

Ok. But why my continuity proof is not a correct one? To prove that its continuous everywhere, we need to choose number c so that the lim (x->c) f(x) = f(c).

[clamtrox]

Consider a closed interval [1+epsilon, 3-epsilon] instead. You should always be able to find a solution in that interval if you make epsilon small enough.

[emyt]

Ok. But why my continuity proof is not a correct one? To prove that its continuous everywhere, we need to choose number c so that the lim (x->c) f(x) = f(c).

yes, but rigourously, you actually have to SHOW that the lim x->c f(x) = f(c). you just can't plug c into f(x) and say that it is therefore continuous