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baldymcgee6
Group Title
For what values of a and b is the following parametric function continuous?
http://screencast.com/t/ydAe0nVzELd
 2 years ago
 2 years ago
baldymcgee6 Group Title
For what values of a and b is the following parametric function continuous? http://screencast.com/t/ydAe0nVzELd
 2 years ago
 2 years ago

This Question is Closed

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
@hartnn
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
I have the solution, but i don't get it... Do you want me to post the solution?
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
http://screencast.com/t/C8XAD3vk1O
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
which part u didn't get ? for limit to be continuous, u need to have lim x>0 ... = lim x>0+ ..
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
why at 0?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
because the point of discontinuity can be 0 only... as the function behaves differently (different definition) before and after 0
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
because of the domain restrictions?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
u can say that, but mostly its because of t<=0 ..... t> 0 if it were t<=2..... t>2, then u would have checked at x=2
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
okay... how does a=b?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
because , for continuity , lim x>0 needs to be equal to x>0+ and lim x>0 was 1 so, x> 0+ = a/b = 1
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
ohhhhh..
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
thank you again!
 2 years ago
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