y = x^2 stationary points: maximum? minimum?

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The stationary points occur when 2x = 0, ie. when x = 0. Is this a maximum or a minimum or what?
 
Re: y = x^2 stationary points

Hello, americo74!

y=x2\displaystyle y\,=\,x^2.
The stationary points occur when 2x = 0, ie. when x = 0.
Is this a maximum or a minimum or what?
Are you familiar with the Second Derivative Test?

The second derivative is: y=2\displaystyle \,y''\,=\,2\, which is always positive.
Hence, the graph is always concave up: \displaystyle \,\cup
Therefore, the stationary point (0,0)\displaystyle (0,0) is a minimum.


A more primitive approach . . .

At x=0\displaystyle x = 0, the derivative is 0 . . . the slope is 0.

To the left at, say, \(\displaystyle x\,=\,-1:\;y'\,=\.-2\) . . . the graph is decreasing: \displaystyle \,\searrow

To the right at, say, x=1:  y=+2\displaystyle x\,=\,1:\;y'\,=\,+2 . . . the graph is increasing: \displaystyle \,\nearrow


Near x=0\displaystyle x\,=\,0, the graph looks like this: \displaystyle \,\searrow\,_{\rightarrow}\,\nearrow

Therefore, the stationary point must be a minimum.
 
That makes sense

I forgot the 2nd derivative test.

However, thank you for reminding me.
 
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