Video: Intermediate Value Theorem

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

02:12

Video Transcript

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

Let’s have a go at sketching this scenario on a graph. We know that 𝑓 of zero is positive. Let’s draw it here and 𝑓 of three is also positive. So maybe, the graph goes through this point and 𝑓 of π‘₯ is continuous. Does this mean that 𝑓 of π‘₯ has no zeros in the interval from zero to three? Well, no. We can sketch a graph of a continuous function 𝑓 for which both 𝑓 of zero and 𝑓 of three are positive, but which has zeros in the closed interval from zero to three. So we’d hope that we can’t use the intermediate value theorem to conclude that 𝑓 of π‘₯ has no zeros because it simply isn’t true.

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

This special case is sometimes known as Bolzano’s theorem. We have to be careful here. The intermediate value theorem does not mean that if 𝑁 is not between 𝑓 of π‘Ž and 𝑓 of 𝑏, then there does not exist a number 𝑐 in the open interval from π‘Ž to 𝑏 such that 𝑓 of 𝑐 equals 𝑁. So the special case does not mean that if 𝑓 of π‘Ž and 𝑓 of 𝑏 have the same sign. In other words, if they’re both positive or both negative, then there cannot exist a number 𝑐, which is a root of 𝑓. This is a statement that we’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 𝑓 of π‘₯ has no zeros in the interval from zero to three.

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