## 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

1. 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.

2. anonymous

Use the quadratic formula to find your roots first:$x=\frac{-2\pm \sqrt{48}}{2}=\sqrt{-2 \pm 4\sqrt{3}}{2}=-1 \pm 2\sqrt{3} \approx -4.464,2.464$

3. anonymous

can you use the quadratic here with the x^3?

4. anonymous

sorry - i misread - i'm off sick today...for a reason it seems

5. anonymous

it's ok, hope you feel better!

6. anonymous

thanks

7. anonymous

the best way is then to just plug the values to find one positive and 1 negative?

8. myininaya

blexthing is right

9. 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).

10. anonymous

ok, got it, thank you all!

11. myininaya

the intermediate value theorem kicks retrice

12. anonymous

welcome

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