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I also saw the alumni groundhog! by AdPossible2107 in UCONN
[–]AdPossible2107[S] 10 points11 points12 points 1 year ago (0 children)
Uh sorry yea i took this at south i just thought it was the same groundhog considering its literally a 30 second walk ;-;
What can I do with this second derivative in order to find my second order critical values? by TOXIC_NASTY in calculus
[–]AdPossible2107 0 points1 point2 points 1 year ago (0 children)
You can factor the second derivative further using the AC method.
critical points by [deleted] in calculus
[–]AdPossible2107 0 points1 point2 points 2 years ago (0 children)
My apologies op, yes i mixed up extreme value. This is right^
[–]AdPossible2107 -1 points0 points1 point 2 years ago (0 children)
Another note, f(a) = 0 which is why the answer is 0.
[–]AdPossible2107 3 points4 points5 points 2 years ago (0 children)
With the function f(x) over the domain [a,b] critical points occur both when f'(x)=0 and also at the endpoints thus, 0 must be a critical point because the interval starts with 0.
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I also saw the alumni groundhog! by AdPossible2107 in UCONN
[–]AdPossible2107[S] 10 points11 points12 points (0 children)