To use the Intermediate Value Theorem to determine which intervals contain at least one zero, we need to find the signs of the function at the endpoints of each interval.
The function f(x) = x + 6x³ - x² - 30x + 4 is a polynomial function.
Let's evaluate f(x) at the endpoints of each interval and determine the signs:
1. Interval (-∞, -4):
The sign of f(-∞) = -∞ + 6(-∞)³ - (-∞)² - 30(-∞) + 4 = -∞ - ∞ + ∞ + ∞ + 4 = -∞
The sign of f(-4) = -4 + 6(-4)³ - (-4)² - 30(-4) + 4 = -4 - 384 - 16 + 120 + 4 = -280
2. Interval (-4, -2):
The sign of f(-4) = -280 (from the previous step)
The sign of f(-2) = -2 + 6(-2)³ - (-2)² - 30(-2) + 4 = -2 - 48 - 4 + 60 + 4 = 10
3. Interval (-2, 0):
The sign of f(-2) = 10 (from the previous step)
The sign of f(0) = 0 + 6(0)³ - (0)² - 30(0) + 4 = 4
4. Interval (0, 4):
The sign of f(0) = 4 (from the previous step)
The sign of f(4) = 4 + 6(4)³ - (4)² - 30(4) + 4 = 4 + 384 - 16 - 120 + 4 = 256
5. Interval (4, ∞):
The sign of f(4) = 256 (from the previous step)
The sign of f(∞) = ∞ + 6(∞)³ - (∞)² - 30(∞) + 4 = ∞ + ∞ - ∞ - ∞ + 4 = ∞
Based on the signs we found, the intervals that have a change in sign and therefore contain at least one zero are:
- Interval (-4, -2)
- Interval (0, 4)
So, the answer is:
(-4, -2), (0, 4)