Math Is Fun Forum

milk242

Member

Registered: 2009-07-09

Posts: 2

Help me solve this problem using Intermediate Value Theorem

Hi everyone,

hope someone can help me... I'm not even sure where to start with this problem...

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

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

Thanks!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Help me solve this problem using Intermediate Value Theorem

Hi milk242

Are you sure about (a / x^3) and (b / x^3) because when a=3 and b=4 there are no roots between (-1,1). Did you mean a*x^3 and b*x^3?

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

milk242

Member

Registered: 2009-07-09

Posts: 2

Re: Help me solve this problem using Intermediate Value Theorem

Hi bobbym,

I'm sure its a/ x^3 and b/x^3

Thanks for your help

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Help me solve this problem using Intermediate Value Theorem

Hi milk242;

  Then you cannot prove that for all a,b that are positive integers there is at least 1 root in (-1,1) because we have 2 immediate counterexamples.

a=3    b=4     x= -1.74164... with 4 more complex roots

a=12  b=12   x= -2               with 4 more complex roots

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Fruityloop

Member

Registered: 2009-05-18

Posts: 143

Re: Help me solve this problem using Intermediate Value Theorem

Bobbym,
     How in the world did you figure that out?  Somehow, you knew to choose 3 and 4 and then figured out the irrational root.
Highly impressive.
     Thank you for your time.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Help me solve this problem using Intermediate Value Theorem

Hi Fruityloop;

Not impressive at all. Try any 2 positive integer values for a and b (I always try small values first when looking for counterexamples). I couldn't find a single pair that yielded a root in (-1,1). Then I just looked for an a and b that yielded a root that was easy to prove by hand, a=12 and b=12 are such numbers.

For getting the root for any choices of a and b I used an automated solver. It also could have been done by hand using interval bisection, the locus and a calculator.

Last edited by bobbym (2009-08-03 07:33:08)

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

riad zaidan

Guest

Re: Help me solve this problem using Intermediate Value Theorem

Hi
I think that the equation is    a/(x^3+2x^2-1) +b/(x^3+x-2) = 0 which is equevalent to 
    a(x^3+x-2)+b(x^3+2x^2-1) = 0 so 
(a+b)x^3 + 2b x^2+ax-2a-b = 0 therefore assume that
f(x)=(a+b)x^3 + 2b x^2+ax-2a-b  we have
1) f is continuous on [-1,1] since it is a polynomial
2) f(-1)=-4a<0 since a>0
    f(1) =2b>0 since b>0
So f satisfies Bolzanoo theoram
tharefore , there exists c in [-1,1] s.t f(c)=0 or satisfies the given equation
w.b.w
Riad Zaidan

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Help me solve this problem using Intermediate Value Theorem

Hi riad zaidan;

Looks OK, there maybe some root in that interval for some specific a and b. But the way he words the problem, though ambiguous, suggests for any a and b that are positive integers that there is a root in that interval. This is not the case. As you can immediately find lots of a's and b's that don't have roots in [-1,1].

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Riad Zaidan

Guest

Re: Help me solve this problem using Intermediate Value Theorem

Hi bobbym ,
If the case as i suggested that a is the numerator for the first expression and also b ,
the statment is true for any +ve real no. a and b, and I dont see any problem.
So please explain!!!!
w.b.w
Riad Zaidan

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Help me solve this problem using Intermediate Value Theorem

Hi Riad Zaidan;

  He says in post #3 it's a/ x^3 and b/ x^3 not a/(x^3+2x^2-1) +b/(x^3+x-2). Then he sounds like he is saying for any positve integer a and b that there is always a root in [-1,1]. This is not true because for  a=3  and  b=4  the roots are x= -1.74164... with 4 more complex roots. So no real roots in [-1,1]. This is a counterexample.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.