# Thread: What does this mean? ( Differentiating — Functions/Equalities )

1. ## What does this mean? ( Differentiating — Functions/Equalities )

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 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?

Thanks so much!!

2. Originally Posted by temhawk
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

...
All your calculations up to this point are correct. But with your problem there is a much easier way:

$-x^2(x^2-4)=0$

A product equals zero if one factor equals zero. Therefore you have:

$-x^2=0~\vee~x^2-4=0~\implies~\boxed{x = 0~\vee~x = -2~\vee~x = 2}$

3. Oooh, thank you very much!

4. 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

5. Originally Posted by temhawk
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 <<<<<< You forgot the negative sign

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
You've made a tiny but very effective mistake. (See above):

$-2x^3+4x = 0~\implies~-2x(x^2-2)=0$

I leave the rest for you.

6. Thanks,

I hate when I forget about the symbols