Question

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

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

it's ok, hope you feel better!

Answer 6

thanks

Answer 7

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

Answer 8

blexthing is right

Answer 9

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).

Answer 10

ok, got it, thank you all!

Answer 11

the intermediate value theorem kicks retrice

Answer 12

welcome