1. ## Number of solutions

Prove that $\displaystyle f(x) = 4x^4+2x^2-6x+2=0$ has exactly two solutions.

Proof so far.

Now $\displaystyle f'(x)=4x^3+4x-6$, set it equals to zero, we have $\displaystyle 2x^3+2x-3=0$, so this equation have 3 solutions, meaning we have 3 extremas, meaning the equation can cross the x-axis twice.

Would that be enough to prove it?

2. $\displaystyle f'(x)=[4x^3+4x-6]'=12x^2+4$
$\displaystyle \forall \mbox{ }12x^2+4>0$
The derivative is always positive, so value of this finction constantly rises and there's just one root (solution).
Is that what you meant?

Prove that $\displaystyle f(x) = 4x^3+4x-6=0$ has exactly two solutions.

Proof so far.

Now $\displaystyle f'(x)=4x^3+4x-6$
No, that is NOT f', it is just f itelf!

, set it equals to zero, we have $\displaystyle 2x^3+2x-3=0$, so this equation have 3 solutions, meaning we have 3 extremas, meaning the equation can cross the x-axis twice.

Would that be enough to prove it?
No, [tex]2x^3+ 2x- 3= 0[/itex] is just your original equation divided by 2. IF it has 3 solutions, then the original certainly does NOT have exactly two solutions! The equation you start with has only one real solution and two complex solutions. Did you miswrite the function?

4. Oh, it should have been $\displaystyle x^4+2x^2-6x+2$, I accidentally put in the derivative instead of the original function.