Using the Intermediate Value Theorem and Rolle's theorem to determine number of roots

Question

Use the Intermediate Value Theorem and Rolle's Theorem to show the that the polynomial $$p(x) = x^{5} + x^{3} + 7x - 2$$ has a unique real root.

Can someone please give some hints on how to do this question. Thanks in advance.

Answer 1

The Intermediate Value Theorem establishes existence: there is at least one real root.

Notice that $p(0) = -2 < 0$ and $p(1) = 7 > 0$. Since $p$ is continuous, the I.V.T. guarantees a number $c$ such that $p(c) = 0$. (In fact, we know that $0 < c < 1$.)

Rolle's Theorem establishes uniqueness: there is at most one real root. Why? Suppose that there were two roots $a, b \in \mathbb{R}$. Since $p$ is differentiable, Rolle's Theorem guarantees a number $c \in (a, b)$ where $p'(c) = 0$. What's wrong with that? The derivative $$ p'(x) = 5x^4 + 3x^2 + 7 > 0 $$ for all $x \in \mathbb{R}$. Why? It's quadratic in $x^2$ and its discriminant $(3)^2 - 4(5)(7) < 0$.

Answer 2

First, $p\,'(x)=5x^4+3x^2+7$; this is a quadratic in the variable $x^2$, so you can find all of its integer roots. Rolle’s theorem says that if $p(a)=0$ and $p(b)=0$, where $a<b$, then there is an $x$ between $a$ and $b$ such that $p\,'(x)=0$. Thus, if $p(x)$ has more than one real root, $p\,'(x)$ has at least one real root. Finally, remember that if $p(a)<0$ and $p(b)>0$, then the intermediate value theorem insures that there is an $x$ between $a$ and $b$ such that $p(x)=0$. That’s everything that you need to use for the problem.

Answer 3

We have $p(0)\lt 0$ and $p(1)\gt 0$, so by the Intermediate Value Theorem there is an $r$ between $0$ and $1$ such that $p(r)=0$.

We now show there is only root. For suppose that $a$ and $b$ are two distinct roots. Then by Rolle's Theorem, there is a $c$ between $a$ and $b$ such that $p'(c)=0$. However. $p'(x)=5x^4+3x^2+7$, so $p'(x)\gt 0$ for all real $x$.