Can we say there is exactly one root of $x^4-7x^3+9=0$ in $(1, \:\: 2)$

calculuspolynomialsquarticsrolles-theoremroots

We have $f(1) \gt 0$ and $f(2) \lt 0$. Hence, the intermediate value theorem (IVT) guarantees at least one root in $(1, \: \: 2)$. Now, let's assume there are two roots $\alpha$ and $\beta$ in $(1, \:\:2 )$. By Rolle's theorem, we have $ \gamma \in (\alpha, \:\: \beta)$ such that $$f'(\gamma)=0$$ we get

$$\gamma =5.25$$ which is not in $(\alpha, \:\: \beta)$. Hence There is exactly one root in $(1, \:\: 2)$. Once Rolle's theorem gives $\gamma$ outside the working interval and at the same time when $IVT$ has guaranteed a root, can we conclude exactly one root in that particular interval?

Best Answer

"Once Rolle's theorem gives γ outside the working interval and at the same time when IVT has guaranteed a root, can we conclude exactly one root in that particular interval?"

No, we can not.

Consider $f(x) = (x-4)(x-1)(x+1)(x+4)= x^4 -17x^2 + 16$ and how many roots it has in the interval $(-2, 5)$.

$f(-2)= (-6)*(-3)*(-1)*4 < 0$ and $f(5) > 0$ so so there is at least one root.

Rolles theorem says that if there are other roots in the interval than

$f'(x) = 4x^3 - 34x=0$ will have have roots in the interval.

And $\gamma = -\sqrt{\frac {17}2}$ is a root of $f'(x) = 0$

and $-\sqrt{\frac {17}2}$ is outside the interval.

But we can't conclude there are no other roots in the interval because although $\gamma $ is outside the interval, $\gamma$ is not the only root of the derivative and there are $\delta = 0$ and $\beta = \sqrt{\frac {17}2}$ that are in the interval.

It's only if there are no roots of the derivative inside the interval that this will work. And one root of the derivative outside the interval does not mean that all roots of the derivative are outside the interval.

So

If there is at least one root, $f(x) = 0$ in an interval $(a,b)$ and if there are no extrema $f'(x) = 0$ inside the interval $(a,b)$, then you can conclude that there is exactly one root in the interval.

but proving there is an extrema outside the interval is not enough. You must prove there is none inside.

.....

On the other hand. Having the roots of the derivative inside the interval doesn't guarantee multiple roots in the interval either.

consider $f(x) = (x-1)x(x+1)+ 6=x^3 - x + 6=0$ and the interval $(-3,-1)$. It has a single root at $x= -2$. It has extrema at $f'(x) = 3x^2 - 1=0; x =\sqrt{\frac 13}$ in the interval. But at both these extrema $f(x) > 0$ so there is no root between them.

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