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ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1To find critical points, you have to differentiate h. Then solve h'(x)=0. The solutions are the critical points.

ame
 2 years ago
Best ResponseYou've already chosen the best response.0HI , i did and sort of stuck at the end 3x(63x^2)^1/2=0, x=0, and (63x^2)^(1/2)=0, so how do i do at the second part... thanks for the help.

dpaInc
 2 years ago
Best ResponseYou've already chosen the best response.0can you retype the function? as written, this is what you have: h(x)= root over(63x^2)

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1OK, so you've got your derivative! Solving h'(x) = 0: \[\frac{ 3x }{ \sqrt{63x^2} }=0 \Leftrightarrow 3x=0\] So x=0

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1Remember: If a/b=0 this means: a=0 (and b is not 0). So there is an easy solution here...

ame
 2 years ago
Best ResponseYou've already chosen the best response.0so, there is only one critical number (0)..

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, we can get no more from this one...

ame
 2 years ago
Best ResponseYou've already chosen the best response.0so what is critical point is sqrt 6 !

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1Here is the graph, it confirms what you found:

ame
 2 years ago
Best ResponseYou've already chosen the best response.0its webwork solution and it is not getting right..

ame
 2 years ago
Best ResponseYou've already chosen the best response.0what would be the critical point ?

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1There is even a maximum in that point.

ame
 2 years ago
Best ResponseYou've already chosen the best response.0the answer (0, sqrt(6)) cameout wrong in the webwork..

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1But the real question now is: are you convinced that it is true?
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