Home Mathematics Prove that 5x^3+4x-2=0 has exactly one solution Prove that 5x^3+4x-2=0 has exactly one solutionMathematics JunehamOctober 23, 20213 CommentsProve that 5x^3+4x-2=0 has exactly one solution
[tex]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 [/tex]The derivative of [tex]5x^3+4x-2[/tex] 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.Reply
step-by-step explanation:
6xy and -16xy000000000
[tex]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 [/tex]
The derivative of [tex]5x^3+4x-2[/tex] 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.