Real analysis question: show that x^4 - x - 1 = 0 has two real roots

[frenchkiki]

Homework Statement

Let f(x)=x^4 - x - 1. Show that f(x)=0 has two real roots.

Homework Equations

None

The Attempt at a Solution

x(x^3 - 1 - 1/x) = 0 which gives x=0 and x^3 - 1 - 1/x=0, x^2 - 1/x - 1/x^2=0, but WolframAlpha says x~~0.724492 and x~~-1.22074. I kept dividing by x it but couldn't come up with a sound result.

 

[gb7nash]

I don't think the point of this problem is to find the roots, but just to show they exist. Are you allowed to use the intermediate value theorem?

 

[ehild]

Find out the behaviour of the function as x goes either to infinity and to minus infinity. It is also useful to calculate f(0), f(1), f(-1), so as you get a picture of the function. Does it have minima and maxima? how many?

ehild

 

[frenchkiki]

I don't think the point of this problem is to find the roots, but just to show they exist. Are you allowed to use the intermediate value theorem?

Yes I am allowed to use the intermediate value theorem.

Find out the behaviour of the function as x goes either to infinity and to minus infinity. It is also useful to calculate f(0), f(1), f(-1), so as you get a picture of the function. Does it have minima and maxima? how many?

as tends to inf: lim f(x) = inf
as tends to -inf: lim f(x) = -inf
f(0)=-1
f(1)=-1
f(-1)=1

I'm not sure on how to find the extremum though

 

[Mark44]

Yes I am allowed to use the intermediate value theorem.



as tends to inf: lim f(x) = inf
as tends to -inf: lim f(x) = -inf

No, this isn't true. As x approaches -∞, f(x) approaches +∞.

f(0)=-1
f(1)=-1
f(-1)=1

I'm not sure on how to find the extremum though

The problem is not asking for this. Focus on what the problem is asking you to show. It might be helpful to sketch a rough graph of this function based on the two limits and the three function values above. Keep in mind the suggestion from gb7nash.

 

[Deveno]

prove that: if f(x) = x⁴ - x - 1,

a) f tends to ∞ as x tends to -∞ or ∞.
b) f'(x) has at least one root
c) f has a global minimum
d) this minimum is < 0.

why does this tell you f has at least 2 real roots?

then show f has at most 2 roots (hint: look at part (b) again).

 

[ehild]

as tends to inf: lim f(x) = inf
as tends to -inf: lim f(x) = -inf
f(0)=-1
f(1)=-1
f(-1)=1

I'm not sure on how to find the extremum though

You see that the function is positive at x=-1, then negative at x=0. As it is continuous, it must be zero somewhere between -1 and 0. It is -1 again at x=-1, so there is a turning point, a minimum between x=0 and x=1. And the function is positive for large x values. Is it zero again somewhere between x=1 and infinity?

ehild

 

[frenchkiki]

OK I figured this one out thanks to you guys:

from the I.V. theorem, there exists c=0 belonging to [a, b] with a=1 and b=-1 in the first case, second case with a=1 and b=2. therefore f(a)=-1<c=0<f(b)=1 and f(a)=-1<c=0<f(2)=13 so we have two real values for x.

Thanks y'all for your help!