Number of solutions

**tttcomrader** · November 13th 2008, 06:43 AM

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

Proof so far.

Now $f'(x)=4x^3+4x-6$ , set it equals to zero, we have $2x^3+2x-3=0$ , 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?

**Arch_Stanton** · November 13th 2008, 07:21 AM

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

**HallsofIvy** · November 13th 2008, 07:34 AM

Originally Posted by tttcomrader

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

Proof so far.

Now $f'(x)=4x^3+4x-6$

No, that is NOT f', it is just f itelf!

, set it equals to zero, we have $2x^3+2x-3=0$ , 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?

**tttcomrader** · November 13th 2008, 07:36 AM

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

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