Here's an idea by factoring out the highest even power and completing the square. Continue with the remaining powers. I'm not sure if actually work, but give it a try:Prove that [math]\forall x\in\R\,\,(3x^6+4x^5-6x^3+3>0)[/math]
Any hint please? I don't need the solution but some hints to prove it! Thanks
Me too. No algebraic short-cut jumps out at me.It's actually not that bad. You know that x = 0 gives a minimum, so you can compare f(0) to 0. Once you find the other minimum value, compare f(value) to 0. I grant that there may be a more formal way to do this proof, but this is the approach I would take.
Don't be. I am working on an answer that I think will work with just a bit of calculus.It's interesting that this problem comes from high school lvl where students don't know Newtons method, I'm confused...
I hope you will get it, I'm trying for some hours...Don't be. I am working on an answer that I think will work with just a bit of calculus.
[math]y(a) = 3x^6+4x^5-6x^3+3 = (3x^6- 6x^3) + (4x^5 + 3) = a(x) + b(x)[/math]Prove that [math]\forall x\in\R\,\,(3x^6+4x^5-6x^3+3>0)[/math]
Any hint please? I don't need the solution but some hints to prove it! Thanks
Can't see the way that it can help ;/I haven't had time to read JeffM's post, but I'll share this thought in case it helps. If you rewrite the inequality as follows then you should be able to prove it for the cases x=0 and x>0 but I can't see an easy way for x<0.
[math]3(x^3 - 1)^2 > -4x^5[/math]
Can you prove the inequality for x=0 and x>0?Can't see the way that it can help ;/
Sir can you prove for x<0?Can you prove the inequality for x=0 and x>0?
[math]\text{For}\, x<0, -4x^5>0 \Rightarrow 3(x^3-1)^2+(-4x^5)>0[/math]I haven't had time to read JeffM's post, but I'll share this thought in case it helps. If you rewrite the inequality as follows then you should be able to prove it for the cases x=0 and x>0 but I can't see an easy way for x<0.
[math]3(x^3 - 1)^2 > -4x^5[/math]
That's good reasoning, what about x>0 then? In case of x=0 is obvious[math]\text{For}\, x<0, -4x^5>0 \Rightarrow 3(x^3-1)^2+4x^5>0[/math]
Sorry it's not a good reasoning...That's good reasoning, what about x>0 then? In case of x=0 is obvious
Sorry missed the negative sign.So
Sorry it's not a good reasoning...
[math]\text{For}\, x<0, -4x^5>0 \Rightarrow 3(x^3-1)^2+(-4x^5)>0[/math]