Proove there is an x s.t. x^3+x=6

[jrsweet]

Homework Statement

Show that there is an x∈R such that x^3+x=6.

The Attempt at a Solution

I'm not exactly sure where to get started with this proof. I think I would need to define a set S={x∈R: x>0 and x^3+x≤6}. Assume S is bounded, and then find lub(S)...?

 

[jgens]

I think with some work, your approach will definitely yield a correct proof; however, are you familiar with the Intermediate Value Theorem?

 

[Gregg]

There are 3 solutions, if you prove one is complex then one must be real?

 

[jrsweet]

How exactly would I use the Intermediate Value Theorem?

 

[jrsweet]

Would I need to first prove that it is continuous then?

 

[Elucidus]

How exactly would I use the Intermediate Value Theorem?

Would I need to first prove that it is continuous then?

All polynomial functions are continuous over all of R.

The Intermediate Value Theorem is used frequently to establish that a solution to an equation exists (but oftern is no help in finding that solution, except by numeric approximation methods).

As an example: Show there is a solution to the equation cos(x) = x.

This is equivalent to showing that cos(x) - x = 0 is solvable.

Since f(x) = cos(x) - x is a continuous function and f(0) = 1 > 0 and f(pi/2) = -pi/2 < 0 there must exist some value c between 0 and pi/2 where f(c) = 0 (by the Intermediate Value Theorem), hence there is a solution to cos(x) = x.

Your problem can be solved similarly.

--Elucidus

 

[njama]

And why don't you simply solve x³+x=6 and show that there are real values for x.

$$x^3+x-6=0$$

Start by checking if $\pm 1, \pm 2, \pm 3, \pm 6$ are solutions of the cubic equation.

 

[HallsofIvy]

And why don't you simply solve x³+x=6 and show that there are real values for x.

$$x^3+x-6=0$$

Start by checking if $\pm 1, \pm 2, \pm 3, \pm 6$ are solutions of the cubic equation.

Well, unless there is an integer root, that won't "solve" the equation! But that is a good way to start. In fact, it is exactly what jgens suggested. If x= 0 $x^3+ x= 0+ 0= 0< 6$. If x= 2, $x^3+ x= 8+ 2= 10> 6$. What does that tell you?

 

[njama]

Well, unless there is an integer root, that won't "solve" the equation! But that is a good way to start. In fact, it is exactly what jgens suggested. If x= 0 $x^3+ x= 0+ 0= 0< 6$. If x= 2, $x^3+ x= 8+ 2= 10> 6$. What does that tell you?

It tells me that the solution is between 0 and 2, i.e $x\in (0,2)$.

Infact if you choose x=1, $1^3+1=1+1=2<6$, so $x \in (1,2)$

If you go by checking for x=1.5 , $1.5^3+1.5=4.875<6$ you will come up with the solution. Now $x \in (1.5 , 2)$ and so on...

Edit: In addition, I came up with better solution.

Check the sign of the discriminant of the cubic equation:

$$D=\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}$$

The general cubic equation is:

$$x^3+px+q$$

Just extract p and q from your equation.

1)If D>0 then there are 1 real and two conjugate complex roots.

2)If D=0 there is one prime and two equal real roots.

3)If D<0 there are 3 real and different roots.