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

1. Nov 5, 2011

### frenchkiki

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

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

2. Relevant equations

None

3. 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.

Last edited: Nov 5, 2011
2. Nov 5, 2011

### 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?

3. Nov 5, 2011

### 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

4. Nov 8, 2011

### frenchkiki

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

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

5. Nov 8, 2011

### Staff: Mentor

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

Last edited: Nov 8, 2011
6. Nov 8, 2011

### 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).

7. Nov 8, 2011

### ehild

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

8. Nov 9, 2011

### 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!