G
Guest
Guest
The stationary points occur when 2x = 0, ie. when x = 0. Is this a maximum or a minimum or what?
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Re: y = x^2 stationary points
Hello, americo74!
The second derivative is: \(\displaystyle \,y''\,=\,2\,\) which is always positive.
Hence, the graph is always concave up: \(\displaystyle \,\cup\)
Therefore, the stationary point \(\displaystyle (0,0)\) is a minimum.
A more primitive approach . . .
At \(\displaystyle x = 0\), 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, \(\displaystyle x\,=\,1:\;y'\,=\,+2\) . . . the graph is increasing: \(\displaystyle \,\nearrow\)
Near \(\displaystyle x\,=\,0\), the graph looks like this: \(\displaystyle \,\searrow\,_{\rightarrow}\,\nearrow\)
Therefore, the stationary point must be a minimum.
Hello, americo74!
Are you familiar with the Second Derivative Test?\(\displaystyle y\,=\,x^2\).
The stationary points occur when 2x = 0, ie. when x = 0.
Is this a maximum or a minimum or what?
The second derivative is: \(\displaystyle \,y''\,=\,2\,\) which is always positive.
Hence, the graph is always concave up: \(\displaystyle \,\cup\)
Therefore, the stationary point \(\displaystyle (0,0)\) is a minimum.
A more primitive approach . . .
At \(\displaystyle x = 0\), 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, \(\displaystyle x\,=\,1:\;y'\,=\,+2\) . . . the graph is increasing: \(\displaystyle \,\nearrow\)
Near \(\displaystyle x\,=\,0\), the graph looks like this: \(\displaystyle \,\searrow\,_{\rightarrow}\,\nearrow\)
Therefore, the stationary point must be a minimum.
G
Guest
Guest
That makes sense
I forgot the 2nd derivative test.
However, thank you for reminding me.
I forgot the 2nd derivative test.
However, thank you for reminding me.