-x^3+8x-12=0 use the theorem on bounds

Question

Use the theorem on bounds to establish the best intergral bounds for the roots of the equation.

 

_x^3+8x-12=0

Answer 1

This can also be written: x³-8x+12=0.

This is a cubic equation so it must have at least one real root. We need to find two values of x such that one integer value makes the expression positive while the other makes it negative. These are integer bounds and the root lies somewhere in between. When x=0, the expression x³-8x+12=12, which is positive. It remains positive for positive integer values of x because x³+12 is always greater than 8x.

So we need to try x<0. When x=-1, the expression=19; when x=-2, it's 20, so the expression is decreasing in value. At x=-3 it drops to 9. At x=-4 it's -20, so now we have two integer values which produce a change in sign: x=-3 and x=-4. The root lies in between these two integers.