Question

Amistre pls ,if u can, answer my q:
y=sqrx Find min(using 1st derivative)
Is it 0?

Answer 1

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

Answer 2

the min of the function defined by y = sqrt(x) across its domain would be 0

Answer 3

\[[\sqrt{x}]'=\frac{1}{2\sqrt{x}}\]
there is no 0 of this derivative, but it is "undefined" at 0; so it is a point of interest to look into

Answer 4

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.