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

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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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.