use the intermediate value theorem to choose an interval over which function f(x)=-2x^3-3x+5 is guaranteed to have a zero.

[-3,-2]
[-2,0]
[0,2]
[2,4]

To use the Intermediate Value Theorem to select an interval where the function f(x) = -2x^3 - 3x + 5 is guaranteed to have a zero, we need to check for a change in sign of the function values within the interval.

Let's evaluate the function at the endpoints of each interval and see if there is a sign change between the function values:

1. Interval [-3, -2]:
- f(-3) = -2(-3)^3 - 3(-3) + 5 = -2(-27) + 9 + 5 = 54 + 9 + 5 = 68
- f(-2) = -2(-2)^3 - 3(-2) + 5 = -2(-8) + 6 + 5 = 16 + 6 + 5 = 27

Since f(-3) = 68 > 0 and f(-2) = 27 > 0, there is no sign change in this interval.

2. Interval [-2, 0]:
- f(-2) = -2(-2)^3 - 3(-2) + 5 = -2(-8) + 6 + 5 = 16 + 6 + 5 = 27
- f(0) = -2(0)^3 - 3(0) + 5 = 0 + 0 + 5 = 5

Since f(-2) = 27 > 0 and f(0) = 5 > 0, there is no sign change in this interval.

3. Interval [0, 2]:
- f(0) = -2(0)^3 - 3(0) + 5 = 0 + 0 + 5 = 5
- f(2) = -2(2)^3 - 3(2) + 5 = -2(8) - 6 + 5 = -16 - 6 + 5 = -17

Since f(0) = 5 > 0 and f(2) = -17 < 0, there is a sign change in this interval. Hence, the Intermediate Value Theorem guarantees that the function has a zero within this interval.

4. Interval [2, 4]:
- f(2) = -2(2)^3 - 3(2) + 5 = -2(8) - 6 + 5 = -16 - 6 + 5 = -17
- f(4) = -2(4)^3 - 3(4) + 5 = -2(64) - 12 + 5 = -128 - 12 + 5 = -135

Since f(2) = -17 < 0 and f(4) = -135 < 0, there is no sign change in this interval.

From our analysis, the only interval that guarantees the existence of a zero for the function f(x) = -2x^3 - 3x + 5 is [0, 2].