The Intermediate Value Theorem (IVT) states that if a continuous function takes on two values at two different points, then it must take on every value between those two values at least once in that interval.
To apply this theorem to the equation 2x³ + x² – 2 = 0, we first define the function:
f(x) = 2x³ + x² – 2
Next, we evaluate this function at the endpoints of the interval [1, 2].
1. Calculate f(1):
f(1) = 2(1)³ + (1)² – 2 = 2 + 1 – 2 = 1
2. Calculate f(2):
f(2) = 2(2)³ + (2)² – 2 = 16 + 4 – 2 = 18
Now, we have f(1) = 1 and f(2) = 18. Both values are positive, so we check if there is a value within the interval where the function might cross the x-axis (where it equals zero).
Additionally, to find a value where the function is negative, we can check a point within the interval, like f(0):
f(0) = 2(0)³ + (0)² – 2 = -2
Now we have:
- f(0) = -2 (negative)
- f(1) = 1 (positive)
Since f(0) is negative and f(1) is positive, by the Intermediate Value Theorem, there must be at least one value c in the interval (0, 1) where f(c) = 0.
Thus, we can conclude that there is a root of the equation 2x³ + x² – 2 = 0 in the interval [1, 2].