WEBVTT
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In this video, we’ll learn about the intermediate value theorem, a very intuitive but powerful theorem about continuous functions.
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But before we talk about this theorem, let’s first talk about the square root function.
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What is the square root of four?
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Well, this one is easy.
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We know it’s two.
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And we can check that this is true by squaring two, multiplying it by itself.
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And doing so, we do indeed get four.
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Well, negative two squared is also four.
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So perhaps, there’s no argument for saying that the square root of four is negative two as well.
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But as we’d like square roots to be a function, we need a unique answer for the square root of four.
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And we pick the positive one.
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That’s two.
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How about the square root of 49 over 25?
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This is slightly trickier.
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But we can verify that the answer is seven over five by squaring seven over five and seeing that we get 49 over 25.
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To find the square root of 10.89, you might have to reach for our calculators.
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But having got the value of 3.3, it’s easy to check.
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We can even check it by hand if we don’t trust our calculators.
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But what about the square root of two?
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Your calculator might tell you that its value is 1.414213562 or something like that.
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And if you square this value using your calculator, you may well get the answer two, which is good news, right?
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However, when I square this using my computer, which is more accurate than my calculator, I find that actually, this number squared is 1.99999999894, which is very close to, but not equal to, two.
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The reason this happens is that this value 1.414213562 is not exact.
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The decimal expansion continues 370395 and it just keeps going.
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You might know that the square root of two is an irrational number.
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And so, we can’t write it as a fraction like seven over five.
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And so, we can’t verify that its square really is two in the same way that we verified that the square of seven over five really was 49 over 25.
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Nor does the decimal expansion of the square root of two terminate, the digits just keep coming.
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So we can’t square it using a long multiplication, as we did for 3.3.
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In fact, we can’t even write down its exact value, at least not as a fraction or decimal.
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If we’re asked to write it exactly, we just write the square root of two, which seems like cheating somehow.
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The square root of two equals the square root of two.
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Nevertheless, we call the square root of two a real number.
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And we say the same for the square root of three, which has the same issues.
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On the other hand, we say that the square root of negative one is not a real number.
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Why this discrimination?
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The answer comes from the intermediate value theorem.
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Here is a graph of 𝑦 equals 𝑥 squared.
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We can square numbers using this graph.
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For example, we can find one squared by identifying one on the 𝑥-axis, going up from one on the 𝑥-axis to where it intersects the curve and then along to the 𝑦-axis.
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And we read off one squared, which is one.
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We can also read off square root using this graph.
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For example, to find the square root of four, we identify four on the 𝑦-axis and go in the other direction along to the curve and then down until we hit the 𝑥-axis at two.
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So the square root of four is two.
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Now to find the square root of two, we can go along from two on the 𝑦-axis until we hit the curve.
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And then going down to the 𝑥-axis gives us the square root of two.
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If we call this number, 𝑐, then we must have that 𝑐 squared equals two.
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So 𝑐 is the square root of two.
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Similarly, we can find the square root of three.
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Will this allow us to find the square root of any number?
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Well, not quite, for a negative number will miss the curve entirely.
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We can’t find the square root of negative one in this way.
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We could find the square roots of two and three because these values lie between the values of 𝑓 of one, which is one, and 𝑓 of two, which is four.
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The function is continuous on the interval from one to two.
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And so, the 𝑦-value on the curve, which is the value of the function, changes smoothly from one to four, passing through two and three as it goes.
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The curve must, therefore, intersect 𝑦 equals two and 𝑦 equals three.
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And furthermore, the 𝑥-values at which these intersections occur must lie in the interval from one to two.
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We can write down in words why we think this is true.
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Let 𝑓 of 𝑥 equals 𝑥 squared as two is between 𝑓 of one, which is one, and 𝑓 of two, which is four.
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And as 𝑓 is continuous, there must exist a number 𝑐 which lies in the open interval from one to two, such that 𝑓 of 𝑐 equals two.
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In other words, 𝑐 squared is two and 𝑐 is the square root of two.
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Of course, it doesn’t just work for the square root of two.
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We can also find the square root of three in the same way.
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What about five?
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Well, five does not lie between 𝑓 of one which is one and 𝑓 of two, which is four.
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But it does lie between 𝑓 of two, which is four, and 𝑓 of three, which is nine.
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And so, there must exist 𝑐 in the open interval from two to three such that 𝑓 of 𝑐 is five.
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For any positive real number 𝑁, we can find 𝑎 and 𝑏 such that 𝑁 is between 𝑓 of 𝑎 and 𝑓 of 𝑏.
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And so, its square root 𝑐 must lie between 𝑎 and 𝑏.
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Here, we’ve been talking about the function 𝑓 of 𝑥 equals 𝑥 squared and finding square root using this.
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But we could just as easily be talking about cube roots, the same applies or fifth roots.
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In fact, the only thing we use about 𝑓 is that it is continuous.
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We can, therefore, state a very general result.
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If 𝑓 is continuous on the closed interval from 𝑎 to 𝑏 and the number 𝑁 lies between 𝑓 of 𝑎 and 𝑓 of 𝑏, then there exists 𝑐 in the open interval from 𝑎 to 𝑏, such that 𝑓 of 𝑐 equals 𝑁.
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You can check that this is the same statement we have above, just with 𝑓 of 𝑥 being continuous.
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Let’s draw a more general graph to illustrate this.
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We have some function 𝑓, which is continuous on the closed interval 𝑎, 𝑏 and 𝑁 is between 𝑓 of 𝑎 and 𝑓 of 𝑏.
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Then, there exists 𝑐 in the open interval from 𝑎 to 𝑏 such that 𝑓 of 𝑐 equals 𝑁.
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This appears to be true.
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Note that depending on the value of 𝑁, this choice of 𝑐 may not be unique.
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There might be many possible values of 𝑐, but we’re guaranteed at least one.
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We need 𝑁 to be between 𝑓 of 𝑎 and 𝑓 of 𝑏 for this to hold.
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In other words, we need 𝑁 to be an intermediate value of 𝑓 of 𝑎 and 𝑓 of 𝑏.
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If 𝑁 isn’t an intermediate value of 𝑓 of 𝑎 and 𝑓 of 𝑏, then we might get some values of 𝑐.
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But there are no guarantees.
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For this reason, the statement is called the intermediate value theorem.
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It’s a theorem because despite the fact that it may seem obvious, it’s actually something that can and needs to be proved using the definition of continuity.
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However, the proof is quite technical, using the technical definition of continuity.
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And so, we won’t be seeing it in this video.
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What we will be seeing are plenty of applications.
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Let’s see our first.
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The figure shows the graph of the function 𝑓 on the closed interval from zero to 16 together with the dashed line with equation 𝑦 equals 30.
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𝑓 of zero is less than 30 and 𝑓 of 16 is greater than 30.
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But 𝑓 of 𝑥 is not equal to 30 anywhere on the closed interval from zero to 16.
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Why does this not violate the intermediate value theorem?
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Let’s just verify what we’re told in the question.
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Is 𝑓 of zero less than 30?
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Well, yes, we can see here 𝑓 of zero appears to be about 12.
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And likewise, 𝑓 of 16 is greater than 30.
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It appears to be about 32.
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But 𝑓 of 𝑥 is not equal to 30 anywhere on the interval.
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This is true because the dashed line 𝑦 equals 30 nowhere intersects the graph of our function.
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The question is, why does this not to violate the intermediate value theorem?
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Well, what is the intermediate value theorem?
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It states that if a function 𝑓 is continuous on a closed interval from 𝑎 to 𝑏 and if the number 𝑁 is between 𝑓 of 𝑎 and 𝑓 of 𝑏, the values of the function on the end points of the interval.
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Then there exists 𝑐 in the open interval from 𝑎 to 𝑏 such that 𝑓 of 𝑐 is 𝑁.
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We can see then how it might appear that we have a counterexample to the intermediate value theorem.
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We set 𝑁 equal to 30 and note that 30 is between 𝑓 of zero and 𝑓 of 16.
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However, there doesn’t exist any value 𝑐 in the open interval from zero to 16 such that 𝑓 of 𝑐 is 30.
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Why is this not a counterexample to the intermediate value theorem?
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Well, the intermediate value theorem only applies if 𝑓 is continuous.
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Our function doesn’t satisfy this required hypothesis.
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We can see that there is a discontinuity here at 𝑥 equals eight.
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So why does this not violate the intermediate value theorem?
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Because the function is not continuous at 𝑥 equals eight.
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And so, it is not continuous on the closed interval from zero to 16 which would be required for the intermediate value theorem to apply.
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Now that we’ve seen that we can’t apply the intermediate value theorem to functions which are not continuous, let’s see why the intermediate value theorem is useful for functions which are.
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Let 𝑓 of 𝑥 equal three to the 𝑥 power minus 𝑥 to the fifth power.
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According to the intermediate value theorem, which of the following intervals must contain a solution to 𝑓 of 𝑥 equals zero?
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Is it the closed interval from two to three, the closed interval from zero to one, the closed interval from negative three to negative two, the closed interval from one to two, or the closed interval from negative two to negative one?
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So we have a function.
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How can we use the intermediate value theorem to tell which of these intervals has a root of this function, a solution to 𝑓 of 𝑥 equals zero?
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Well, let’s remind ourselves of the intermediate value theorem.
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It states that if a function 𝑓 is continuous on the closed interval from 𝑎 to 𝑏 and 𝑁 is some number between the values of the function at the end points of that interval.
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That’s 𝑓 of 𝑎 and 𝑓 of 𝑏.
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Then there exists some number 𝑐 in the open interval from 𝑎 to 𝑏, such that 𝑓 of 𝑐 equals 𝑁.
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The first thing to notice that our function 𝑓 is continuous on the real numbers and so will be continuous on any of the intervals in the options as well.
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So this hypothesis holds.
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Now remember, we want to find a solution to 𝑓 of 𝑥 equals zero.
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Comparing this with 𝑓 of 𝑐 equals 𝑁, it looks like we want to set 𝑁 equal to zero.
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So the intermediate value theorem is telling us that for our continuous function 𝑓, if zero is between 𝑓 of 𝑎 and 𝑓 of 𝑏, then there exists 𝑐 in the open interval from 𝑎 to 𝑏 such that 𝑓 of 𝑐 is zero.
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In other words, if 𝑓 of 𝑎 and 𝑓 of 𝑏 have different signs, then there is some number 𝑐 between 𝑎 and 𝑏 which is a root of 𝑓.
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So to solve this question, we take each interval in the options in turn, starting with the interval from two to three.
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And if the signs of 𝑓 of two and 𝑓 of three are different, if one of them is positive and one of them is negative, then we know there must be a root, a solution to 𝑓 of 𝑥 equal zero in this interval.
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So let’s compute 𝑓 of two and 𝑓 of three.
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We do this using the definition for 𝑓 of 𝑥 we have in the question.
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We can evaluate these by hand or using a calculator, finding that 𝑓 of three is negative 216 and 𝑓 of two is negative 23.
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There’s no sign change of the function here.
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Both values are negative.
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By the intermediate value theorem, we know that the function 𝑓 must take all values between negative the 23 and negative 216 as its input changes from two to three.
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So we’d have a solution to 𝑓 of 𝑥 equals negative 100.
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For example, in this interval.
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However, as zero is not between negative 23 and negative 216, we can’t say that there must be a solution to 𝑓 of 𝑥 equals zero in this interval.
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We move on to option B.
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The closed interval from zero to one.
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We compute the values of the function at the end points.
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We find that 𝑓 of one is two and 𝑓 of zero is one.
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Again, there’s no sign change, so no guarantee of a zero in this interval.
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However, we can see a sign change between 𝑓 of one and 𝑓 of two.
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𝑓 of one is positive and 𝑓 of two is negative.
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The intermediate value theorem tells us that as 𝑓 is continuous on the closed interval from one to two and as zero is between 𝑓 of one, which is two, and 𝑓 of two, which is negative 23.
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Then there exists a number 𝑐 in the open interval from one to two such that 𝑓 of 𝑐 is zero.
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And as 𝑐 lies in the open interval from one to two, it must also lie in the closed interval from one to two.
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And so we have a solution to 𝑓 of 𝑥 equals zero in the closed interval from one to two.
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This is option D.
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We can — if we like — check the values of the function at the end point of the other intervals in the options to see that there is no sign change in either intervals C or E.
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And so, D is definitely the only correct answer.
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While the intermediate value theorem guarantees us a root or solution to 𝑓 of 𝑥 equals zero in the interval from one to two, we can’t say just based off the intermediate value theorem that there are no roots in the other intervals.
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Let’s see why not?
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If 𝑓 of 𝑥 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 𝑥 has no zeros in the interval from zero to three?
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Let’s have a go at sketching this scenario on a graph.
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We know that 𝑓 of zero is positive.
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Let’s draw it here and 𝑓 of three is also positive.
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So maybe, the graph goes through this point and 𝑓 of 𝑥 is continuous.
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Does this mean that 𝑓 of 𝑥 has no zeros in the interval from zero to three?
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Well, no.
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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.
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So we’d hope that we can’t use the intermediate value theorem to conclude that 𝑓 of 𝑥 has no zeros because it simply isn’t true.
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Why might we think that the intermediate value theorem implies this incorrect statement?
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What the intermediate value theorem states is that if 𝑓 is continuous on the closed interval from 𝑎 to 𝑏 and 𝑁 is the number between 𝑓 of 𝑎 and 𝑓 of 𝑏, then there exists a number 𝑐 in the open interval from 𝑎 to 𝑏 such that 𝑓 of 𝑐 equals 𝑁.
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Setting 𝑁 equal to zero, we get the special case that if 𝑓 is a continuous function and 𝑓 of 𝑎 and 𝑓 of 𝑏 have opposite signs, then there exists a number 𝑐 in the open interval from 𝑎 to 𝑏 such that 𝑓 of 𝑐 equals zero.
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In other words, there is a zero of the function 𝑓 in the interval.
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This special case is sometimes known as Bolzano’s theorem.
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We have to be careful here.
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The intermediate value theorem does not mean that if 𝑁 is not between 𝑓 of 𝑎 and 𝑓 of 𝑏, then there does not exist a number 𝑐 in the open interval from 𝑎 to 𝑏 such that 𝑓 of 𝑐 equals 𝑁.
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So the special case does not mean that if 𝑓 of 𝑎 and 𝑓 of 𝑏 have the same sign.
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In other words, if they’re both positive or both negative, then there cannot exist a number 𝑐, which is a root of 𝑓.
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This is a statement that we’re asked about in the question and it does not follow from the intermediate value theorem.
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Our answer is, therefore, no.
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We cannot conclude that 𝑓 of 𝑥 has no zeros in the interval from zero to three.
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Let’s now reflect on the key points we’ve covered in this video.
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First, the statement of intermediate value theorem.
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If 𝑓 is continuous on the closed interval 𝑎𝑏 and 𝑁 is some number between 𝑓 of 𝑎 and 𝑓 of 𝑏, then there exists 𝑐 between 𝑎 and 𝑏 for which 𝑓 of 𝑐 equals 𝑁.
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This may seem obvious to you, but not everything that’s obvious is actually true.
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However, this is a theorem that can be proved.
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It’s important to interpret the statement of the theorem correctly.
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The theorem doesn’t say that 𝑐 must be unique.
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There may be more than one value of 𝑐 in the open interval for which 𝑓 of 𝑐 is 𝑁.
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Nor does the theorem claim that if 𝑚 is not between 𝑓 of 𝑎 and 𝑓 of 𝑏, then there is no 𝑑 for which 𝑓 of 𝑑 equals 𝑚.
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This just simply isn’t true as we can see by looking at the diagram.
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We’ve also seen that the intermediate value theorem can be used to justify the existence of zeros of functions.
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You might not think that this is particularly useful.
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But it turns out to be a very powerful tool that can be used to prove things which aren’t at all obvious.