Prove using Rolle's Theorem that an equation has exactly one real solution.

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.

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 $x\ne0$.

...

Integrate the function once, $$g(x)=\int x^7+x^5+x^3+1~dx$$ In the interval $[-1,0]$ observe that, $g(-1)\neq g(0)$. Hence by Rolle's theorem there does not exist any $x\in (-1,0)$ such that $g'(x)=f(x)=0$. Hence there are no roots between $[-1,0]$. I see no other way but to differentiate $$f(x)=x^7+x^5+x^3+1$$ to get $$f^{'}(x)=7x^6+5x^4+3x^2$$ Hence $f^{'}\geq 0$ which means it is always increasing.

$f(-\infty)=-\infty$ and $f(\infty)=\infty$ . Hence the curve will cut the x-axis at only one point. But its difficult to tell in my opinion at which point it will cut.

More generally, let $f(x) =\sum a_i x^{m_i} $ where the $m_i$ are positive odd integers and each $a_i > 0$. Then I claim that $f(x)$ has exactly one real root and this root is negative.

Proof: Since all $a_i > 0$, $f(x) > 0$ for $x > 0$. Also, $f(x) < 0$ for $x < \min(-1, 1\big/\sum a_i) $, so $f$ has a negative real root.

Finally, since $f'(x) =\sum m_ia_i x^{m_i-1} $ has all positive coefficients and even exponents, $f'(x) > 0$ for all real $x$, so $f$ can have at most one real root, and, therefore, has exactly one real root.

Most of problems I see with your proof aren't mathematical. It's just far too wordy and awkward to read. Below, I made a few changes which make it much more readable.

Since f(x) = x^7+x^5+x^3+1 is a polynomial then it is continuous over all the real numbers, (-∞,∞). Then it is certainly continuous on [-1,0] and the Intermediate Value Theorem applies. The I.V.T. states that since f(x) is continuous on [-1,0], for any number C between f(-1)=-2 and f(0)=1, then there exists a number c with -1 < c < 0 such that f(c)=C. If $C = 0$, we see there exists at least one solution.

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:[a,b]->R is continuous on [a,b], differentiable on (a,b), and satisfies f(a)=f(b), then there exists a point c ϵ (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). Then f'(c)=0 for some $c \in (a,b)$.

At this point in your proof, you have to convince me why $f'(c) = 0$ is impossible, using the function itself.