Question

How do you use the intermediate value theorem to show that there is a root of the equation `2x^3+x^2+2=0` over the interval (-2, -1)?

Answer 1

First find the y values of the ends of the interval so that the function is easier to visualize:
Let `f(x)= 2x^3+x^2+2`

`f(-2)=2(-2)^3+(-2)^2+2`
`=-16+4+2=-10`

`f(-1)=2(-1)^3+(-1)^2+2`
`=-2+1+2=1`

IVT states that if a continuous function f(x) on the interval [a,b] has values of opposite sign inside an interval, then there must be some value x=c on the interval (a,b) for which f(c)=0.

Because f(-2) is negative and f(-1) is positive, and f(x) is continuous on the closed interval [-2,-1], there must be some value x=c on the interval [-2,-1] for which f(c)=0. f(x) is continuous on the interval [-2,-1] because it is a polynomial cubic function, and is continuous at each point in the interval.