### Video Transcript

A ball is thrown into the air. Its vertical position above the ground is given by the function π₯ of π‘ is equal to negative three π‘ squared plus 12π‘ plus seven. According to the intermediate value theorem, during which of the following intervals must the ball land on the ground? A) The closed interval from four to five, B) the closed interval from one to two, C) the closed interval from two to three, D) the closed interval from three to four, or E) the closed interval from five to six.

We can start by noticing that the question tells us that our function π₯ of π‘ represents the vertical position of the ball above the ground. So, in particular, when our function π₯ of π‘ is equal to zero, the ball is exactly on the ground. This means that the question is asking us to use the intermediate value theorem to find the interval which the ball must land on the ground, which is equivalent to using the intermediate value theorem to find the interval which the function π₯ of π‘ is equal to zero. So we can ask the question, What does the intermediate value theorem say?

We recall that the intermediate value theorem states that if a function π of π₯ is continuous on the closed interval from π to π and π is a number which is between π evaluated at π and π evaluated at π, then there must exist a number π in the open interval from π to π such that π evaluated at π is equal to π. The question is asking us to find an interval on which the ball must have landed on the ground, which we know is equivalent to finding an interval on which the function π₯ of π‘ is evaluated at some point to be equal to zero. What this means in our intermediate value theorem is that we want to set our function to be equal to π₯ of π‘ and we want π to be equal to zero.

Substituting this information into intermediate value theorem would then give us that π₯ of π‘ is continuous on the closed interval from π to π and zero is between π₯ evaluated at π and π₯ evaluated at π. Then, we can conclude that there must exist a π in the open interval from π to π, such that π₯ evaluated at π is equal to zero. This would then let us conclude that the ball landed on the ground when π‘ is equal to π. So, in particular, we could conclude that the ball landed on the ground at some point in the open interval from π to π.

There are two requirements to use an intermediate value theorem. We need to find an π and a π such that zero is between π₯ evaluated at π and π₯ evaluated at π. And we also need to show that our function π₯ of π‘ is continuous on the closed interval from π to π. Letβs start by considering the continuity of our function π₯ of π‘. We know that π₯ of π‘ is equal to negative three π‘ squared plus 12π‘ plus seven, which is a polynomial. And we also know that all polynomials are continuous on the set of real numbers. Therefore, since our function π₯ of π‘ is a polynomial, we could conclude that π₯ of π‘ is continuous on the real numbers. And in particular, for any real numbers π and π, π₯ of π‘ will be continuous on the closed interval from π to π.

So we have shown that the first requirement for using the intermediate value theorem is true. Now, we need to find the values of π and π such that zero is between π₯ evaluated at π and π₯ evaluated at π. There are many different ways to find these values of π and π. Since weβre working with a quadratic, weβre going to find the roots of this quadratic. We can find the roots for quadratic equation by using the quadratic formula, giving us roots at negative 12 plus or minus the square root of 12 squared minus four multiplied by negative three multiplied by seven all divided by two multiplied by negative three, which we can calculate to give us that the roots of π₯ of π‘ are at π‘ is approximately equal to negative 0.5 and π‘ is approximately equal to 4.5.

If we were then to sketch a graph of our function π₯ of π‘, we would get a negative quadratic where the roots are approximately negative 0.5 and 4.5. From our sketch, we can then see that when π‘ is between zero and 4.5, our function is above the axis. This means that the outputs of π₯ are greater than zero when π‘ is between zero and 4.5. Similarly, we can see when the values of π‘ go above our root at approximately 4.5, the function goes below the axis. So the outputs of π₯ of π‘ are less than zero when π‘ is greater than 4.5.

What this means is if we choose π to be between zero and 4.5 and π to be greater than 4.5, then our function π₯ evaluated at π must be greater than zero and our function π₯ evaluated at π must be less than zero. So as an example, we could choose π to be equal to four and π to be equal to five. And we would know that π₯ evaluated at four would be greater than zero and π₯ evaluated at five would be less than zero.

Just to be sure, we can calculate these values. π₯ evaluated at four is equal to negative three multiplied by four squared plus 12 multiplied by four plus seven, which we can calculate to be equal to seven. Similarly, we can evaluate π₯ at five to be equal to negative three multiplied by five squared plus 12 multiplied by five plus seven, which if we calculate, we will see that π₯ evaluated at five is equal to negative eight.

What this means is that when we set π in the intermediate value theorem to be equal to four and π to be equal to five, then zero is between π₯ evaluated at four and π₯ evaluated at five. So the second requirement of our intermediate value theorem is also true. Therefore, since we have shown that both requirements for the intermediate value theorem are true, we can conclude that there exists a π in the open interval from four to five such that π₯ evaluated at π is equal to zero.

Finally, since we know that there exists a π in the open set from four to five, we can extend this to say that π could also be in the closed interval from four to five. Therefore, we can conclude that the ball must have touched the ground during the interval from four to five, which was our option A.