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

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- Jul 29th 2009, 05:32 AMJoelFind 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 - Jul 29th 2009, 05:38 AMsongoku
4(x^3 - x) = 0

4x(x^2 - 1) = 0

Can you continue? ^^ - Jul 29th 2009, 05:53 AMHallsofIvy
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 __________.