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!
Best Answer
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.