Using Rolle's Theorem to prove roots.

Question

Show that $x^5+10x+3=0$ has exactly one real solution using Rolle's Theorem.

I am referencing a closely related stack answer here. So I have tried letting $y=x^5+10x+3$. Then $y'=5x^4+10$. Set $y'=0$ $$0=5x^4+10$$ $f$ is continuous on $\mathbb{R}$ and $f$ is differentiable on $\mathbb{R}$ but I need a closed interval $[a.b]$ of continuity that corresponds to an open interval $(a,b)$ of differentiation. Where $f(a)=f(b)$.

I do not see any real solution such as they were able to find in the provided link. As well as I do not see how to satisfy $f(a)=f(b)$. Any help and explanation is appreciated!

Answer 1

Assume $f(x)=x^5+10x+3$ has two solutions $f(a)=f(b)=0, \space a<b$, then according to Rolle's theorem $\exists c \in (a,b): f'(c)=0$. But $f'(x)=5x^4+10>0$ and has no solutions, contradiction. At this point, we know that $f(x)$ has $1$ or $0$ solutions.

Using Intermediate value theorem $f(-1)=-8$ and $f(0)=3$, thus $\exists c \in (-1,0): f(c)=0$.

Combining these two facts, $f(x)$ has exactly one solution.

Answer 2

$f'(x)=5x^4+10\geq 10>0$ for all real $x.$ So $f:\mathbb R\to \mathbb R$ is one-to-one.

Because if $f(a)=f(b)$ with $a<b$ then the function $g(x)=f(x)-f(a)=f(x)-f(b)$ is $0$ at $x=a$ and at $x=b$, implying $g'(x)=0$ for some $x\in (a,b)$ by Rolle's theorem. But $g'(x)=f'(x)\ne 0$ for all $x$.

Therefore there is at most one real $x$ with $f(x)=0.$

Since $f(0)>0$ and $f(-2)<0$ and $f$ is continuous we have $0\in [f(-2),f(0)]\subset \{f(x):x\in [-2,0]\} $.

Therefore there is at least one real $x$ with $f(x)=0. $

Answer 3

Slight variation with fewer calculations: After you use Rolle's theorem, it suffices to note that a root exists, since $$ \lim_{x\rightarrow \infty}f(x)=+\infty $$ and $$ \lim_{x\rightarrow -\infty}f(x)=-\infty $$ Since polynomials are continuous, there is at least one root.

Note: This shows any odd degree polynomial has a real root!