Solving x^7-x+17=0: Can I Make My Own Interval?

[mateomy]

If I am given a problem that asks of me:

What is the largest number of real roots that $x^7-x+17=0$ can have? Be sure to show why your answer is correct.

But it doesn't give an interval to work with, can I just make up my own interval, say [0,3]? Because I know this is continuous and differentiable everywhere.

 

[lanedance]

i would guess you should consider the whole real line

 

[mateomy]

Okay, so how could I show that numerically? Would I just have to show that it is differentiable and continuous by doing the derivation? And as far as maximum zeros, would it just be (Aaaahhhhh, I can't remember the theorem's name...) Descartes(?) theory on polynomial roots, which is just one integer less the highest exponent, so 6 possible real roots?

 

[lanedance]

how about considering the derivative... in an interval where the derivative is always one sign, there can only be at most one axis crossing...

 

[mateomy]

So then...

$$f'(x)7x^6-1$$

would mean that it can only cross the x-axis once because there is only 'one' sign. Would that be right?

 

[lanedance]

Okay, so how could I show that numerically? Would I just have to show that it is differentiable and continuous by doing the derivation? And as far as maximum zeros, would it just be (Aaaahhhhh, I can't remember the theorem's name...) Descartes(?) theory on polynomial roots, which is just one integer less the highest exponent, so 6 possible real roots?

have a look at this
http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra
so if you include the multiplicity, every polynomial of order n, has n roots in complex space.

so by that reasoning you could have up to 7 roots.

However we require them to be real & distinct, so based on the location of turning points of the function, it should be easy to read off how many axis crossing there are...

 

[mateomy]

I see. So is my previous post accurate then? (sorry if this is annoying, just straightening out my knowledge).

 

[lanedance]

So then...

$$f'(x)=7x^6-1$$

would mean that it can only cross the x-axis once because there is only 'one' sign. Would that be right?

nope, i see three regions of different signs...

finding the turning points should be pretty instructive

 

[mateomy]

So you're taking the sign changes off of the initial function not its derivative then. Thats the only place I see three sign changes. As opposed to the two in the derivative.

 

[lanedance]

you can consider both, but mainly the function itself - for the function to cross the axis it must change from +ve to -ve, however between the critical points you know the function is acting monotonically

now for the derivative, there will be 2 turning points, so 2 sign changes, and 3 regions of different sign

 

[mateomy]

Okay, you couldn't possibly have made it clearer. And stated that way, I can picture it graphically in my head. Thanks for putting up with my questions...off to more studying.

 

[HallsofIvy]

By the way, DesCarte's "rule of signs",which predates Calculus, and so Rolle's theorem, says that the number of positive roots of a polynomial equation is at most equal to the number of sign changes of the coefficients as you go down from the highest to the lowest power. $x^7- x+ 17= 0$ has two sign changes (+1 to -1 to +17) and so has at most 2 positive roots. If we change -x for x we get $-x^7+ x+ 17= 0$ which has only one sign change and so at most one positive root. But a positive root of $-x^7+ x+ 17= 0$ is a negative root of $x^7- x+ 17= 0$ so that equation has at most 3 real roots.