I have to get the null-point(s) of a curve/function. The original function goes like this:
(i tried to use the MATH BBCode tags but I don't know how to use them so they caused errors)
f(x) = -x⁴ / 2 + 2x²
To get the null-points (not sure if it's called that in English) I have to set the above to be equal to zero:
-x⁴ / 2 + 2x² = 0
Next, I do ×2 on the equation:
-x⁴ + 4x² = 0
Then I say that x² = u so that I can use a formula later
-u² + 4u = 0
Now I use this formula to get u₁ and u₂
(I think it's from Vieta)
x² + px + q = 0
₁x₂ = -p / 2 ± √((p / 2)² - q)
In my equation I don't have a q so I don't know if that conflicts with the proper use of the formula. Anyway, I went on to replace the variables in the formula with the ones from my equation (leaving out the - q part inside the square-root section):
₁u₂ = -2 ± √4
That leaves me with these results for ₁u₂
u₁ = 0
u₂ = -4
Since u is actually x² we get ₁x₂ from the square-root of u₁
x₁ = 0
x₂ = 0
Now my first question is if the above is correct. Is x₂ zero as well, if u₁ was zero? ( ₁x₂ = ±√u₁ )
Next, I'd do the same with u₂ but since u₂ is -4 and you can't get the square-root of negative numbers (without resorting to Complex Numbers) this leaves me with no answer to x₃ and x₄
Is this true, or what am I doing/understanding wrong?
Please help me out quickly, I need this homework done today!
Thanks so much!!
I'm stuck again.
This time I have to get the extreme values. So I set the first derivative of x of the function to be equal to zero.
f(x) = -x⁴ / 2 + 2x²
f'(x) = 4x³ / 2 + 4x = 2x³ + 4x
2x³ + 4x = 0
I continued the same way as before:
2x (x² + 2) = 0
So x₁ = 0 but I can't get x₂ or x₃