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 $\displaystyle u= x^2$. Then $\displaystyle y= x^4- 2x^2+ 3= u^2- 2u+ 3$. You can find the minimum value of that by completing the square: $\displaystyle u^2- 2u+ 1+ 2= (u- 1)^2+ 2$
The mimimum value of $\displaystyle u^2- 2u+ 3$ is 2 at u= 1 so the minimum value of $\displaystyle x^4- 2x^2+ 3$ is __________.