# Number of solutions

• Nov 13th 2008, 06:43 AM
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?
• Nov 13th 2008, 07:21 AM
Arch_Stanton
\$\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?
• Nov 13th 2008, 07:34 AM
HallsofIvy
Quote:

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!

Quote:

, 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?
• Nov 13th 2008, 07:36 AM