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For what values of k is f(x) continuous?
3x^211x4 x<=4
kx^22x1 x>4
 one year ago
 one year ago
For what values of k is f(x) continuous? 3x^211x4 x<=4 kx^22x1 x>4
 one year ago
 one year ago

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bahrom7893Best ResponseYou've already chosen the best response.1
when x = 4, f(x) must be 0
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.1
there's an alternative method...
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.1
which is kinda what i did, but i skipped steps
 one year ago

bahrom7893Best 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
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.1
Thus the second equation must also be 0 when x=4
 one year ago

bahrom7893Best 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
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.1
So kx^22x1=0 when x=4
 one year ago

acrossBest ResponseYou've already chosen the best response.1
You want\[\lim_{x\to4}\left(kx^22x1\right)=0.\]
 one year ago

lizlozadaBest ResponseYou've already chosen the best response.0
so the answer k= 9/16, that's it?
 one year ago

acrossBest 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.
 one year ago

lizlozadaBest ResponseYou've already chosen the best response.0
ok, what's the process? plugging in the 4?
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.1
it's looking at where the function "breaks"... in this case it's at 4
 one year ago

bahrom7893Best 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
 one year ago

bahrom7893Best 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
 one year ago

lizlozadaBest ResponseYou've already chosen the best response.0
they will always give you what you have to plug in?
 one year ago

acrossBest ResponseYou've already chosen the best response.1
I'll ask an interesting question: is the function differentiable at that point? ;)
 one year ago

lizlozadaBest ResponseYou've already chosen the best response.0
we haven't learned what that means...
 one year ago
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