Answer
1. Use the Intermediate Value Theorem with $f(-1)=-2$ and $f(1)=4$ to show that at least one zero exists in $[-1,1].$
2. Use Rolle's theorem to show that there can not be more zeros on that interval.
3. Generalize the finding to any interval $[-a,a]$.
$f$ has exactly one real solution.
Work Step by Step
$f(x)=x^{5}+x^{3}+x+1$
$f^{\prime}(x)=5x^{4}+3x^{2}+1$
$f$ is differentiable for all $x.$
Note that $f^{\prime}(x)$ has even powers of x, leading to $f^{\prime}(x) > 0$
1.
Because $f(-1)=-2$ and $f(1)=4$, (function values have different signs)
by the Intermediate Value Theorem , $f$ has at least one zero, $c$, in $[-1,\ 1], .$
2.
Let us ASSUME that $f$ had more than one, say 2 zeros, $f(c_{1})=f(c_{2})=0$.
Then, by Rolle's Theorem
there exists of a number $c_{3}\in(c_{1},c_{2})$ such that
$f^{\prime}(c_{3})=f(c_{2})-f(c_{1})=0.$
But, $f^{\prime}(x) > 0$ for all $x$. There is no $c_{3 }$ such that $f^{\prime}(c_{3})=0.$
We have a contraditcion.
Our ASSUMPTION was wrong.
So, $f$ has exactly one zero in $[-1,\ 1]$.
3.
We now generalize and take any interval $[-a,a]$, where $a > 0.$
$f(-a)$ is negative, $f(a)$ is positive,
Applying the same logic as above, there is only one zero in $[-a,a]$.
Since a can be arbitrarily large,
$f$ has exactly one real solution on R.
