anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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.
anonymous
  • anonymous
Use the quadratic formula to find your roots first:\[x=\frac{-2\pm \sqrt{48}}{2}=\sqrt{-2 \pm 4\sqrt{3}}{2}=-1 \pm 2\sqrt{3} \approx -4.464,2.464\]
anonymous
  • anonymous
can you use the quadratic here with the x^3?

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anonymous
  • anonymous
sorry - i misread - i'm off sick today...for a reason it seems
anonymous
  • anonymous
it's ok, hope you feel better!
anonymous
  • anonymous
thanks
anonymous
  • anonymous
the best way is then to just plug the values to find one positive and 1 negative?
myininaya
  • myininaya
blexthing is right
anonymous
  • anonymous
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).
anonymous
  • anonymous
ok, got it, thank you all!
myininaya
  • myininaya
the intermediate value theorem kicks retrice
anonymous
  • anonymous
welcome

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