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lizlozada
Group Title
For what values of k is f(x) continuous?
3x^211x4 x<=4
kx^22x1 x>4
 2 years ago
 2 years ago
lizlozada Group Title
For what values of k is f(x) continuous? 3x^211x4 x<=4 kx^22x1 x>4
 2 years ago
 2 years ago

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bahrom7893 Group TitleBest ResponseYou've already chosen the best response.1
when x = 4, f(x) must be 0
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.1
k*4^22*41=0
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.1
there's an alternative method...
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.1
which is kinda what i did, but i skipped steps
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.1
3x^211x4, when x<=4.. plug in 4 into that equation: 3*4^211*44=3*16444=48444=0
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.1
Thus the second equation must also be 0 when x=4
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.1
dw:1348118755000:dw lol btw that's not what the functions look like, but you have to make them touch each other
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.1
So kx^22x1=0 when x=4
 2 years ago

across Group TitleBest ResponseYou've already chosen the best response.1
You want\[\lim_{x\to4}\left(kx^22x1\right)=0.\]
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.1
16k81=0 k=9/16
 2 years ago

lizlozada Group TitleBest ResponseYou've already chosen the best response.0
so the answer k= 9/16, that's it?
 2 years ago

across Group TitleBest ResponseYou've already chosen the best response.1
@bahrom7893 is right. That's the only value of \(k\) for which the piecewise defined function is continuous.
 2 years ago

lizlozada Group TitleBest ResponseYou've already chosen the best response.0
ok, what's the process? plugging in the 4?
 2 years ago

lizlozada Group TitleBest ResponseYou've already chosen the best response.0
then solving for k?
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.1
it's looking at where the function "breaks"... in this case it's at 4
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.1
Plug that part into the known piece of the function.. so you know that y is some number, in our case 0
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.1
So in order for our function not to break, the second piece must start where the first piece left off, or at 0
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.1
when x = 4
 2 years ago

lizlozada Group TitleBest ResponseYou've already chosen the best response.0
they will always give you what you have to plug in?
 2 years ago

across Group TitleBest ResponseYou've already chosen the best response.1
I'll ask an interesting question: is the function differentiable at that point? ;)
 2 years ago

lizlozada Group TitleBest ResponseYou've already chosen the best response.0
we haven't learned what that means...
 2 years ago
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