# Thread: Find a minimum value

1. ## Find a minimum value

Hi all,

I am stuck at the end of this.

Find the minimum value of the curve y = x⁴ - 2x² + 3 in the domain -3 ≤ x ≤ 3.

y = x⁴ - 2x² + 3
y’ = 4x³ - 4x
stationary points where y’ = 0
4x³ - 4x = 0
4(x³ - x) = 0
Stuck here

2. 4(x^3 - x) = 0
4x(x^2 - 1) = 0

Can you continue? ^^

3. Originally Posted by Joel
Hi all,

I am stuck at the end of this.

Find the minimum value of the curve y = x⁴ - 2x² + 3 in the domain -3 ≤ x ≤ 3.

y = x⁴ - 2x² + 3
y’ = 4x³ - 4x
stationary points where y’ = 0
4x³ - 4x = 0
4(x³ - x) = 0
Stuck here
Let $u= x^2$. Then $y= x^4- 2x^2+ 3= u^2- 2u+ 3$. You can find the minimum value of that by completing the square: $u^2- 2u+ 1+ 2= (u- 1)^2+ 2$

The mimimum value of $u^2- 2u+ 3$ is 2 at u= 1 so the minimum value of $x^4- 2x^2+ 3$ is __________.