Question

Prove that 5x^3+4x-2=0 has exactly one solution

Answer 1

$5x^3+4x-2=0\\\\ (5x^3+4x-2)'=15x^2+4\\\\ 15x^2+4=0\\ 15x^2=-4\\ x^2=-\dfrac{4}{15}\\ x\in\emptyset$

The derivative of $5x^3+4x-2$ is positive for all real numbers, which means the function is increasing in its all domain and therefore there is only one intersection with the x-axis ⇒ one solution.