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Amistre pls ,if u can, answer my q: y=sqrx Find min(using 1st derivative) Is it 0?
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the min is found at 'a' zero of the first derivative yes; but just becasue we get f' = 0 doesnt gaurentee a min or max
the min of the function defined by y = sqrt(x) across its domain would be 0
\[[\sqrt{x}]'=\frac{1}{2\sqrt{x}}\] there is no 0 of this derivative, but it is "undefined" at 0; so it is a point of interest to look into
That's why I asked...cause 0 can't be a value of y' but it can b of y and from the graph of y=sqrx we have a min at (0;0) see the attach.
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