### Video Transcript

True or False: The extreme value theorem states that only continuous functions on closed bounded intervals have maxima and minima.

In this question, we’re given a statement about the extreme value theorem, and we need to determine whether the statement is true or false. And there are surprisingly several different ways of answering this question. The easiest way would just be to recall exactly what the extreme value theorem states. However, before we do this, let’s analyze the statement the question is giving us.

The statement tells us that only continuous functions on closed bounded intervals will have maxima and minima. And there are many different conditions for this statement to hold. For example, since this statement says that this will only hold for continuous functions, if we showed this was true for a noncontinuous function, the statement would be false.

Similarly, we could also check for a continuous function on an open interval or a continuous function on an unbounded interval. And in fact, these conditions lead us to many counterexamples to the statement. And we can go through a few of these now. First, consider the constant function of one. And we can sketch the graph of this function on the coordinate plane. It’s a horizontal line of 𝑦 is equal to one. Since this is a constant function, we know it’s a continuous function. However, instead of considering this function on a closed bounded interval, we’ll consider this function on the entire set of real numbers, which is not a bounded set.

We can see in the definition of the function or from its graph that this has maxima and minima. The maximum value of this function is one and the minimum value is also one because all outputs of the function are one. So in particular, our function has a maximum value and it has a minimum value.

However, the statement we’re given in the question says that only continuous functions on closed bounded intervals will have this property. We’ve given a counterexample to this. This is a function on a nonbounded interval with this property. Therefore, we can show that the statement in the question is not a theorem because there’s a counterexample to the statement. We didn’t even need to recall the definition of the extreme value theorem.

It’s also worth noting we could’ve found a number of different counterexamples to all of the other conditions. For example, we could consider the same function on the open set from zero to one. This then gives us a continuous function on a bounded interval. However, it’s not a closed interval. And once again, we still have maxima and minima.

Finally, let’s show one more counterexample to the continuous function criterion of the statement.

Consider the following sketch of the piecewise-defined function 𝑦 is equal to 𝑓 of 𝑥, where 𝑓 of 𝑥 is one when 𝑥 is less than or equal to zero and 𝑓 of 𝑥 is equal to negative one when 𝑥 is greater than zero. We can see in the diagram of this function this is not a continuous function. There’s a jump discontinuity at 𝑥 is equal to zero. However, we can see from the diagram the maximum possible output of our function is one and the minimum possible output of our function is negative one. Therefore, this function has maximum and minimum values. However, it’s not a continuous function. So once again, the statement given to us in the question can’t be true.

Let’s now recall exactly what the extreme value theorem tells us. We recall the extreme value theorem tells us if 𝑓 is a real-valued continuous function on a closed interval from 𝑎 to 𝑏, then 𝑓 attains a maximum and a minimum value on the closed interval from 𝑎 to 𝑏. So the extreme value theorem does not give us a property that all functions with maximum and minimum values have. Instead, it’s a property that all continuous functions on closed bounded intervals have. Therefore, we were able to show the statement given to us in the question is false.