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

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

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

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

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

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

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

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

ZeHanz
 one year 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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