# Question Video: Intermediate Value Theorem

If π(π₯) is continuous over [0, 3], π(0) > 0, and π(3) > 0, can we use the intermediate value theorem to conclude that π(π₯) has no zeros in the interval [0, 3]?

02:12

### Video Transcript

If π of π₯ is continuous over the closed interval from zero to three, π of zero is greater than zero, and π of three is greater than zero, can we use the intermediate value theorem to conclude that π of π₯ has no zeros in the interval from zero to three?

Letβs have a go at sketching this scenario on a graph. We know that π of zero is positive. Letβs draw it here and π of three is also positive. So maybe, the graph goes through this point and π of π₯ is continuous. Does this mean that π of π₯ has no zeros in the interval from zero to three? Well, no. We can sketch a graph of a continuous function π for which both π of zero and π of three are positive, but which has zeros in the closed interval from zero to three. So weβd hope that we canβt use the intermediate value theorem to conclude that π of π₯ has no zeros because it simply isnβt true.

Why might we think that the intermediate value theorem implies this incorrect statement? What the intermediate value theorem states is that if π is continuous on the closed interval from π to π and π is the number between π of π and π of π, then there exists a number π in the open interval from π to π such that π of π equals π. Setting π equal to zero, we get the special case that if π is a continuous function and π of π and π of π have opposite signs, then there exists a number π in the open interval from π to π such that π of π equals zero. In other words, there is a zero of the function π in the interval.

This special case is sometimes known as Bolzanoβs theorem. We have to be careful here. The intermediate value theorem does not mean that if π is not between π of π and π of π, then there does not exist a number π in the open interval from π to π such that π of π equals π. So the special case does not mean that if π of π and π of π have the same sign. In other words, if theyβre both positive or both negative, then there cannot exist a number π, which is a root of π. This is a statement that weβre asked about in the question and it does not follow from the intermediate value theorem. Our answer is, therefore, no. We cannot conclude that π of π₯ has no zeros in the interval from zero to three.