Prove that the equation $x^7+x^5+x^3+1=0$ has exactly one real solution. You should use Rolle’s Theorem at some point in the proof.
Since $f(x) = x^7+x^5+x^3+1$ is a polynomial then it is continuous over all the real numbers, $(-\infty,\infty)$. Since $f(x)$ is continuous on $(-\infty,\infty)$ then it is certainly continuous on $[-1,0]$ and the Intermediate Value Theorem (IVT) applies. The IVT states that since $f(x)$ is continuous on $[-1,0]$ we can let $C$ be any number between $f(-1)=-2$ and $f(0)=1$, namely $C=0$, then there exists a number $c$ with $-1 < c < 0$ such that $f(c)=C=0$. Since $f(x)$ is continuous on $[-1,0]$ and differentiable on $(-1,0)$ then Rolle’s Theorem applies. Rolle’s Theorem states that if a function $f\colon [a,b] \to \mathbf{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$ then if $f(a)=f(b)$, there exists a point $c \in (a,b)$ such that $f'(c)=0$. We assume that there is more than one real solution for this equation, namely $f(a)=0=f(b)$. If there exists more than one real solution for $f(x)=0$ then $f(a)=0=f(b)\implies a=b$, and thus there is only one real solution to the equation, as desired.
But I feel I am missing something, and I'm not sure how to show that the equation is differentiable on the interval and that Rolle's Theorem applies.
Best Answer
Let $y = x^7+x^5+x^3+1$
$y(0) = 1, y(-1)=-2$
By IVT, there exists at least one real root in $x\in(-1,0)$ such that $y(x)=0$.
Now I claim that there is EXACTLY one such real root, by using the method of contradiction.
Suppose not, there exists at least $2$ real roots $x_1,x_2$ such that $y(x_1)=0,y(x_2)=0$
Since $y$ is differentiable, by Rolle's theorem, there exists a number $a\in(x_1,x_2)$ such that $y'(a)=0$. However, $y'(x)=7x^6+5x^4+3x^2>0$ for all $x\ne0$.
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