anonymous
  • anonymous
For what values of k is f(x) continuous? 3x^2-11x-4 x<=4 kx^2-2x-1 x>4
Mathematics
jamiebookeater
  • jamiebookeater
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bahrom7893
  • bahrom7893
when x = 4, f(x) must be 0
bahrom7893
  • bahrom7893
k*4^2-2*4-1=0
bahrom7893
  • bahrom7893
there's an alternative method...

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bahrom7893
  • bahrom7893
which is kinda what i did, but i skipped steps
bahrom7893
  • bahrom7893
3x^2-11x-4, when x<=4.. plug in 4 into that equation: 3*4^2-11*4-4=3*16-44-4=48-44-4=0
bahrom7893
  • bahrom7893
Thus the second equation must also be 0 when x=4
bahrom7893
  • bahrom7893
|dw:1348118755000:dw| lol btw that's not what the functions look like, but you have to make them touch each other
bahrom7893
  • bahrom7893
So kx^2-2x-1=0 when x=4
across
  • across
You want\[\lim_{x\to4}\left(kx^2-2x-1\right)=0.\]
bahrom7893
  • bahrom7893
16k-8-1=0 k=9/16
anonymous
  • anonymous
so the answer k= 9/16, that's it?
across
  • across
@bahrom7893 is right. That's the only value of \(k\) for which the piece-wise defined function is continuous.
anonymous
  • anonymous
ok, what's the process? plugging in the 4?
anonymous
  • anonymous
then solving for k?
bahrom7893
  • bahrom7893
it's looking at where the function "breaks"... in this case it's at 4
bahrom7893
  • bahrom7893
Plug that part into the known piece of the function.. so you know that y is some number, in our case 0
bahrom7893
  • bahrom7893
So in order for our function not to break, the second piece must start where the first piece left off, or at 0
bahrom7893
  • bahrom7893
when x = 4
anonymous
  • anonymous
they will always give you what you have to plug in?
across
  • across
I'll ask an interesting question: is the function differentiable at that point? ;)
anonymous
  • anonymous
we haven't learned what that means...

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