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anonymous

  • 5 years ago

the polynomial f(x) = x^3 + 2x - 11 has a real zero between which 2 consecutive integers: a: 0 and 1 b: 1 and 2 c: 2 and 3 d: 3 and 4 e: 4 and 5

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  1. anonymous
    • 5 years ago
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    b. 1 and 2 In order for there to be a real zero the graph most likely went from a y-value that is negative to a y-value that is positive. When you cross from negative to positive you must pass across the x-axis which is a zero. If you put in 1 for all x's - you get out -8 so the point (1,-8) is on the graph. If you put in 2 for all x's - you get out 1 so the point (2,1) is on the graph. If you draw a line between these two points, you will see it crossed the x-axis.

  2. anonymous
    • 5 years ago
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    Use the quadratic formula to find your roots first:\[x=\frac{-2\pm \sqrt{48}}{2}=\sqrt{-2 \pm 4\sqrt{3}}{2}=-1 \pm 2\sqrt{3} \approx -4.464,2.464\]

  3. anonymous
    • 5 years ago
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    can you use the quadratic here with the x^3?

  4. anonymous
    • 5 years ago
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    sorry - i misread - i'm off sick today...for a reason it seems

  5. anonymous
    • 5 years ago
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    it's ok, hope you feel better!

  6. anonymous
    • 5 years ago
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    thanks

  7. anonymous
    • 5 years ago
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    the best way is then to just plug the values to find one positive and 1 negative?

  8. myininaya
    • 5 years ago
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    blexthing is right

  9. anonymous
    • 5 years ago
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    Just plug each pair of numbers into your equation and if you get two function values that change sign, it MUST be the case that the function cut the x-axis (i.e. a root exists in that interval).

  10. anonymous
    • 5 years ago
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    ok, got it, thank you all!

  11. myininaya
    • 5 years ago
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    the intermediate value theorem kicks retrice

  12. anonymous
    • 5 years ago
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    welcome

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