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lizlozada

For what values of k is f(x) continuous? 3x^2-11x-4 x<=4 kx^2-2x-1 x>4

  • one year ago
  • one year ago

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

    • one year ago
  2. bahrom7893
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    k*4^2-2*4-1=0

    • one year ago
  3. bahrom7893
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    there's an alternative method...

    • one year ago
  4. bahrom7893
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    which is kinda what i did, but i skipped steps

    • one year ago
  5. bahrom7893
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    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

    • one year ago
  6. bahrom7893
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    Thus the second equation must also be 0 when x=4

    • one year ago
  7. 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

    • one year ago
  8. bahrom7893
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    So kx^2-2x-1=0 when x=4

    • one year ago
  9. across
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    You want\[\lim_{x\to4}\left(kx^2-2x-1\right)=0.\]

    • one year ago
  10. bahrom7893
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    16k-8-1=0 k=9/16

    • one year ago
  11. lizlozada
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    so the answer k= 9/16, that's it?

    • one year ago
  12. across
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    @bahrom7893 is right. That's the only value of \(k\) for which the piece-wise defined function is continuous.

    • one year ago
  13. lizlozada
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    ok, what's the process? plugging in the 4?

    • one year ago
  14. lizlozada
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    then solving for k?

    • one year ago
  15. bahrom7893
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    it's looking at where the function "breaks"... in this case it's at 4

    • one year ago
  16. 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

    • one year ago
  17. 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

    • one year ago
  18. bahrom7893
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    when x = 4

    • one year ago
  19. lizlozada
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    they will always give you what you have to plug in?

    • one year ago
  20. across
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    I'll ask an interesting question: is the function differentiable at that point? ;)

    • one year ago
  21. lizlozada
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    we haven't learned what that means...

    • one year ago
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