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lizlozada

  • 2 years ago

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

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  1. bahrom7893
    • 2 years ago
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    when x = 4, f(x) must be 0

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

  3. bahrom7893
    • 2 years ago
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    there's an alternative method...

  4. bahrom7893
    • 2 years ago
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    which is kinda what i did, but i skipped steps

  5. bahrom7893
    • 2 years ago
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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

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

  7. bahrom7893
    • 2 years ago
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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

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

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

  10. bahrom7893
    • 2 years ago
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    16k-8-1=0 k=9/16

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

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

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

  14. lizlozada
    • 2 years ago
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    then solving for k?

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

  16. bahrom7893
    • 2 years ago
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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

  17. bahrom7893
    • 2 years ago
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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

  18. bahrom7893
    • 2 years ago
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    when x = 4

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

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

  21. lizlozada
    • 2 years ago
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    we haven't learned what that means...

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