Number of solutions

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November 13th 2008, 06:43 AM
tttcomrader

Number of solutions

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?
November 13th 2008, 07:21 AM
Arch_Stanton

$f'(x)=[4x^3+4x-6]'=12x^2+4$
$\forall \mbox{ }12x^2+4>0$
The derivative is always positive, so value of this finction constantly rises and there's just one root (solution).
Is that what you meant?
November 13th 2008, 07:34 AM
HallsofIvy

Quote:

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!

Quote:

, 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?
November 13th 2008, 07:36 AM
tttcomrader

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