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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?

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    $\begingroup$ Interval between $1$ and $2$ $\endgroup$ – Umesh shankar Jun 29 '18 at 17:27
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Yes, we can. The fact that the equation $f'(x)=0$ has a solution outside the interval doesn't matter. What matters is that it has no root in the interval.

Besides, note that the equation $f'(x)=0$ actually has two roots outside the interval.

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  • $\begingroup$ what matters is that it has no root in the interval? You mean for $f'(x)=0$? $\endgroup$ – Umesh shankar Jun 29 '18 at 17:20
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    $\begingroup$ Yes. That's what I wrote and that's what I mean. $\endgroup$ – José Carlos Santos Jun 29 '18 at 17:21
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    $\begingroup$ Three roots if you count $x = 0$ as a double root. To elaborate on your answer $f'(x) = 4x^3 -21x^2=x^2(4x -21)=0$ has three roots $x = 0; x=0$ and $x=5.25$. None are in the interval so no other root. $\endgroup$ – fleablood Jun 29 '18 at 17:56
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"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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  • $\begingroup$ Thanks a very good explanation which made me perfectly clear in concept. Also Does IVT and Rolle's theorem gives information about multiplicity of roots? $\endgroup$ – Umesh shankar Jun 29 '18 at 18:33
  • $\begingroup$ Not in and of themselves. But if you have the extrema $x_1, ....., x_k$ so that $x_1 < x_2 < ...... < x_k$ (and these are all the extrema) then if $f(x_i)$ and $f(x_{i+1})$ are opposite signs there is exactly one root between them. If $f(x_i)$ and $f(x_{i+1})$ are the same signs then there are no roots between them. $\endgroup$ – fleablood Jun 29 '18 at 18:50
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The posted proof works as long as you show that $f'$ has no roots in $(1,2)$, but for a shortcut...

We have $f(1) \gt 0$ and $f(2) \lt 0$

You can then use Descartes' rule of signs to determine that $f$ has either $2$ positive real roots, or none at all. Given that there is one root in $(1,2)$ the latter case is excluded, and the second positive root must be in $(2,\infty)$ since $f(2) \lt 0$ and $\lim_{x \to \infty} f(x) = + \infty$.

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If you have a change of sign and the function is shown to be monotonic (and continuous) in that interval, then there is exactly one root. (If the function is continuous but not monotonic, then there is certainly an odd number of roots but we don't known how many.)

Montonicity is guaranteed when the first derivative keeps the same sign in the interval.

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