Prove that the equation $x^{15} + 7x^{3} - 5 = 0$ has exactly one real solution 
Prove that the equation $x^{15} + 7x^{3} - 5 = 0$ has exactly one real solution.

A hint that has been given by the teacher is to analyze the function $f(x) = x^{15} + 7x^{3} - 5$.
 A: Hints:
Do what your teacher says, and observe:
(1) The equation is a polynomial one of odd degree and thus it has at least one real solution;
(2) $\;f'(x)=15x^{14}+21x^2\;$ is a non-negative function, thus...
A: If $0\le a<b$ show that $a^{15}<b^{15}$, and similarly $7a^3<7b^3$.
So $a^{15}+7a^3 < b^{15}+7b^3$. So the function $g(x)=x^{15}+7x^3$ is an increasing function. Check that $g(0) = 0$ and $g(1)=8$. So $g$'s values change from $0$  to $8$ when $x$ varies from 0 to 1. Being continuous and increasing there is exactly an $x_0\in(0,1)$ such that $g(x_0)=5$. That $x_o$ is the root for $f(x)$. As $g$ is increasing it cannot take the value 5 again. SO that is the only root. 
A: We will use Descartes's rule of signs to determine the number of positive and negative roots of $x^{15}+7x^3-5$. (Zero is obviously not a root.)


*

*The non-zero coefficients of the polynomial per se change sign exactly once. Therefore, it has exactly one positive root.

*Replacing $x$ with $-x$ we get $-x^{15}-7x^3-5$, which has no sign changes at all. Therefore the original polynomial has no negative roots.


In conclusion, the original polynomial has exactly one real root, and it is positive.
