# Thread: How do you show that x^4-4x+1 has exactly 2 real roots?

1. ## How do you show that x^4-4x+1 has exactly 2 real roots?

Using Intermediate Value Theorem and Rolle's Theorem, is there a way to show that x^4-4x+1 has exactly 2 real roots?

I have discovered that a root exists in the interval (0,1) and (1,2). So there must be at least two real roots to the equation. However, I am having difficulty continuing the proof to show that these are the only 2 real roots.

Any help would be appreciated.

2. $f'(x) = 4x^3 - 4$

Let's find all local maxima and minima by setting f'(x) = 0:
$\begin{array}{rcl} 0 & = & 4x^3 - 4 \\ 0 & = & x^3 - 1 \\ 0 & = & (x - 1)(x^2+x+1) \end{array}$

So: $x - 1 = 0 \ \Rightarrow \ x = 1$ and that is all since $x^2 + x + 1 = 0$ has no real roots. So we know there is only one value such that $f'(x) = 0$

Now for the proof. Assume that there are 3 real roots $a_{1}< a_{2}< a_{3}$ so $f(a_{1}) = f(a_{2}) = f(a_{3})$.

Now see if you can use Rolle's Theorem to finish off the proof, keeping in mind f(x) has only 1 value of x such that f'(x) = 0.

4. If you see the sign of the derivative, you'll see that f is strictly decreasing until 1, then strictly increasing.

The minimum is f(1)=1-4+4=-4 < 0

Therefore there is one and only one value from - infinity to 1 such that f(x)=0 and one and only one value from 1 to + infinity such that f(x)=0

Rolle or IVT can be applied to prove it.

5. Originally Posted by Moo
Rolle or IVT can be applied to prove it.
This is a special case of Rolle's theorem, but I like to use the generalized result:

For any integer n > 0, if f is continuous on the interval [a,b], f is differentiable on the interval (a, b), and f(x)=0 has n solutions in [a, b], then f'(x) = 0 has at least (n-1) solutions in (a, b).

If you were to show n-1=1, then n=2. Therefore, f(x) has 2 solutions in which f(x) = 0.

6. So we know there is only one value such that
Originally Posted by Moo
In fact, there are two.
you lost me with that statement.

7. I'm confused too .. There's only one value of x such that $f'(x) = 0$

To add to my earlier post, note that showing that the quartic has at most 2 real roots doesn't prove that it actually does. For example, you can use the same reasoning to $g(x) = x^4 - 4x + 5$ as I did earlier but this function doesn't have any real roots.

That's why noting $f(1) < 0$ is crucial and this is where IVT comes in.

Aaand .. Moo's method was a lot faster and simpler.

8. Much appreciated guys. Thanks.

9. Originally Posted by skeeter
you lost me with that statement.
Pardon me

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