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Let    y   =   x⁴  -  2x²

 

$y\,=\,x^4-2x^2\\~\\ \frac{dy}{dx}\,=\,\frac{d}{dx}x^4-\frac{d}{dx}2x^2\\~\\ \frac{dy}{dx}\,=\,4x^3-4x$

 

Now let's find what values of  x  make  $\frac{dy}{dx}$  be  0

 

$0\,=\,4x^3-4x\\~\\ 0\,=\,4x(x^2-1)$

 

Set each factor equal to zero and solve for  x

 

$\begin{array} {lcl} 4x=0&\qquad\text{or}\qquad &x^2-1=0\\~\\ x=0&\text{or}&x^2=1\\~\\ &&x=1\quad\text{or}\quad x=-1 \end{array}$

 

These are the  x  values of all the critical points.

 

We can look at a graph to determine which is the minimum:

https://www.desmos.com/calculator/vvl7egtes1

 

We can see that the minumum occurs when  x  =  1  and when  x  =  -1

 

When  x  =  ±1 ,    x²  =  1

 

And when  x²  =  1 ,

 

y   =   x⁴  -  2x²   =   (x²)²  -  2x²   =   (1)² - 2(1)   =   -1

 

So the minumum value of  y  is  -1