How can the Intermediate Value Theorem be used to demonstrate that the equation 2x³ + x² + 2 = 0 has at least one root in the interval between -2 and -1?

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The Intermediate Value Theorem (IVT) is a fundamental concept in calculus that helps us determine the existence of roots within a given interval. To show that the equation 2x³ + x² + 2 = 0 has a root in the interval [-2, -1], we will follow these steps:

  1. Define the function: Let f(x) = 2x³ + x² + 2. We want to check whether there is a value of x in the interval [-2, -1] such that f(x) = 0.
  2. Evaluate the function at the endpoints:
    • Calculate f(-2):

      f(-2) = 2(-2)³ + (-2)² + 2 = 2(-8) + 4 + 2 = -16 + 4 + 2 = -10

    • Calculate f(-1):

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

  3. Check the sign change: From our evaluations, we have f(-2) = -10 (which is negative) and f(-1) = 1 (which is positive). Since the function changes signs between these two points, this suggests that there is at least one root in the interval.
  4. Apply the Intermediate Value Theorem: The Intermediate Value Theorem states that if a function is continuous on a closed interval and takes on values of opposite signs at the endpoints, then there exists at least one c in that interval such that f(c) = 0. Because polynomial functions are continuous everywhere, we can apply the IVT here.

In conclusion, since f(-2) is negative and f(-1) is positive, there exists at least one root of the equation 2x³ + x² + 2 = 0 in the interval [-2, -1].

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