Answer:
Choice 3: [0, 2]
Step-by-step explanation:
The Intermediate Value Theorem states that if we have a continuous function <em>f</em> over the interval [a, b] and k is a number between f(a) and f(b), then there must at least one point <em>c</em> within [a, b] such that f(c)=k.
Then, by the IVT, if the endpoints differ in signs, then we must have a zero within the interval since <em>f</em> must cross the x-axis in order to change signs.
So, we will test the endpoint values for each interval.
We have the function:
![f(x)=-2x^3-2x+5](https://tex.z-dn.net/?f=f%28x%29%3D-2x%5E3-2x%2B5)
Choice 1:
Testing for the endpoints, we get:
![\begin{aligned} f(-3)&=-2(-3)^3-3(-3)+5\\&=68\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20f%28-3%29%26%3D-2%28-3%29%5E3-3%28-3%29%2B5%5C%5C%26%3D68%5Cend%7Baligned%7D)
And:
![\begin{aligned} f(-2)&=-2(-2)^3-3(-2)+5\\&=27\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20f%28-2%29%26%3D-2%28-2%29%5E3-3%28-2%29%2B5%5C%5C%26%3D27%5Cend%7Baligned%7D)
Since both values are positive, we are not guaranteed a zero for the interval [-3, -2].
Choice 2:
Testing endpoints, we get:
![f(-2)=27\text{ and } f(0)=5](https://tex.z-dn.net/?f=f%28-2%29%3D27%5Ctext%7B%20and%20%7D%20f%280%29%3D5)
Again, both values are positive, so we are not guaranteed a zero.
Choice 3:
We get:
![f(0)=5\text{ and } f(2)=-17](https://tex.z-dn.net/?f=f%280%29%3D5%5Ctext%7B%20and%20%7D%20f%282%29%3D-17)
Since the values are of different signs, by the IVT, we are guaranteed a zero by for the interval [0, 2] since the function must cross the x-axis in order to become negative.
So, Choice 3 is correct.
Choice 4:
We get:
![f(2)=-17\text{ and } f(4)=-135](https://tex.z-dn.net/?f=f%282%29%3D-17%5Ctext%7B%20and%20%7D%20f%284%29%3D-135)
Both values are negative, so we are not guaranteed a zero.
Note: We <em>may</em> have a zero for the other three intervals. For instance, for [-3, -2], maybe we went from positive to negative to positive again all within the interval [-3, -2]. However, the only interval that guarantees a zero will be C.