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



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bahrom7893
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when x = 4, f(x) must be 0

bahrom7893
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k*4^22*41=0

bahrom7893
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there's an alternative method...

bahrom7893
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which is kinda what i did, but i skipped steps

bahrom7893
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3x^211x4, when x<=4.. plug in 4 into that equation:
3*4^211*44=3*16444=48444=0

bahrom7893
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Thus the second equation must also be 0 when x=4

bahrom7893
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dw:1348118755000:dw
lol btw that's not what the functions look like, but you have to make them touch each other

bahrom7893
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So
kx^22x1=0 when x=4

across
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You want\[\lim_{x\to4}\left(kx^22x1\right)=0.\]

bahrom7893
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16k81=0
k=9/16

lizlozada
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so the answer k= 9/16, that's it?

across
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@bahrom7893 is right. That's the only value of \(k\) for which the piecewise defined function is continuous.

lizlozada
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ok, what's the process? plugging in the 4?

lizlozada
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then solving for k?

bahrom7893
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it's looking at where the function "breaks"... in this case it's at 4

bahrom7893
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Plug that part into the known piece of the function.. so you know that y is some number, in our case 0

bahrom7893
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So in order for our function not to break, the second piece must start where the first piece left off, or at 0

bahrom7893
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when x = 4

lizlozada
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they will always give you what you have to plug in?

across
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I'll ask an interesting question: is the function differentiable at that point?
;)

lizlozada
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we haven't learned what that means...