Using Rolle's theorem and IVT, show that $x^4-7x^3+9=0$ has exactly $2$ roots.

Question

Using Rolle's Theorem and the Intermediate Value Theorem, show that $x^4-7x^3+9=0$ has exactly two roots.

I know how to prove that this equation has at least two real roots by using IVT, but my problem is how do I use Rolle's theorem to prove that it has at most two real roots? I've used Rolle's theorem to prove a function has one real root, but how do I do it with more than one root.

Answer 1

Set $f(x)=x^4-7x^3+9$. The derivative $$f'(x)=4x^3-21x^2=x^2(4x-21)$$ has the sign of $4x-21$, so $f(x)$ is decreasing up to $\frac{21}{4}$, attains a unique minimum $f\bigl(\frac{21}4\bigr)$ and is finally increasing. By the I.V.T. and the definition of the monotony of a function, the equation has exactly two roots if this minimum is negative, none if it is positive and a double root if the minimum is $0$.

This can be generalised to any quartic polynomial $ax^4+bx^3+c$.

Answer 2

The derivative is $4x^3-21x^2=x^2(4x-21)$ is negative if $x< 21/4$ and positive if $x>21/4$, $f(21/4)<0$ and $lim_{-\infty}f(x)=+\infty$ $lim_{+\infty}f(x)=+\infty$. We deduce that $f$ is strictly decreasing in $(-\infty,21/4)$ and has one root in this interval, $f$ is strictly increasing in $(21/4,+\infty)$ and has one root in this interval.

Answer 3

Between two roots of a polynomial the derivative must vanish at least once, by Rolle’s theorem.

Since the derivative vanishes at $0$ and $21/4$, because $f'(x)=x^2(4x-21)$, there can be as much as three roots.

A refined version of Rolle’s theorem, however, tells us that the derivative must vanish at a local maximum or minimum between two roots; thus we can exclude $0$, where the derivative doesn't change sign (the function is decreasing in a neighborhood of $0$). Hence the roots are at most two.

Their number is even, because clearly the function hasn't multiple roots; so we have either no root or two roots.

Since $$ f(0)=9, \qquad f(6)=-207, \qquad f(7)=9 $$ the IVT tells you that there is a root in $(0,6)$ and a root in $(6,7)$.

Answer 4

Let $f(x)=x^4-7x^3+9$

$D=\mathbb{R}$ ( $f(x)$ is continuos on $\mathbb{R}$)

//Prove $f(x)$ has at least 2 real roots//

We have:

$f(1) = 3 >0$; $f(2) = -31 <0$

$\therefore$ $f(x)$ has at least one real root between $1$ and $2$ (intermediate value theorem) $(1)$

$f(6) = -207 <0$; $f(7)=9>0$

$\therefore$ $f(x)$ has at least one real root between $6$ and $7$ (intermediate value theorem) $(2)$

From $(1)$ and $(2)$ ==> $f(x)$ has at least 2 real roots.

//Prove $f(x)$ has exactly 2 real roots//

Consider $f'(x)=0$ ==> $4x^3-21x^2=0$

<=> $x=0$ $\cup$ $x=\frac{21}{4}$

We have $f(0)=9 >0$ and $f(\frac{21}{4})=-\frac{62523}{256} <0$

==> $f(x)$ cannot has more than $3$ real roots.

However, the second root we've proved earlier (between $6$ and $7$) > $\frac{21}{4}$ (the graph of $f(x)$ is now increasing to infinity)

==> $f(x)$ has exactly 2 real roots ==> QED.

(I recommend you draw a graph on GeoGebra or Desmos to understand clearly)