The Intermediate Value Theorem

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. At six in the morning, the temperature outside is fifteen degrees. By noon, it is thirty degrees. Was there a moment when it was exactly twenty degrees? Of course, right? Temperature does not jump. It changes smoothly, so it must pass through every value in between. Now here is the surprise. This common sense is a theorem, and it can prove that an equation has a root without ever solving it! Even an equation like cos x equals x, which no simple algebra can solve. Questions on where a root lies come up often in JEE, so let's begin.

0:40The theorem: no jumps, so no value is skipped

So, here is the theorem. It is called the intermediate value theorem, or IVT for short. Start with a function f, continuous on the closed interval a, b. The square brackets mean both ends, a and b, are included, okay? And continuous means the graph has no breaks and no jumps anywhere on that interval. With me? Now pick any number k that lies between f of a and f of b. Think of f of a and f of b as the heights of the graph at the two ends. Right? Now picture it. The graph starts at height f of a and ends at height f of b, with no breaks. Can it get from one height to the other without ever touching the height k? No, it cannot! Just like our temperature this morning. So, f of c equals k, for some c in the closed interval a, b. The double arrow means implies, and this curved symbol, like a rounded E, means belongs to. Some c means at least one point where the function takes exactly the value k. Clear?

1:48The special case: a sign change means a root

Now, the form we use the most. Suppose f of a and f of b have opposite signs: one negative, one positive. Okay? Then tell me. Which number lies between a negative number and a positive number, for sure? Zero, exactly! So take k equals zero, and the theorem gives f of c equals zero, for some c. That c is a root! And notice the round brackets now: the open interval a, b, with the ends left out. Why? Neither f of a nor f of b is zero, so the root cannot sit at an end. Got it? Quick check from the course. Show that x cubed minus x minus one equals zero has a root between one and two. The small raised three means x to the power three. What will you compute? Just the signs at the two ends! Call the left side f of x. f of one is one minus one minus one, that is minus one, and minus one is less than zero; that is the less-than sign. f of two is eight minus two minus one, that is five, greater than zero, the greater-than sign. Okay? And f is a polynomial, so it is continuous everywhere. Continuous, and the sign changes, so there is a root in the open interval one, two. That's it! We never solved the equation, and we did not need to.

3:13Careful: no continuity, no theorem

Now, a trap, straight from the course. Take f of x equals one over x, on the closed interval minus one, one. f of minus one is minus one, less than zero, and f of one is one, greater than zero. Opposite signs! So by our theorem, one over x equals zero somewhere between minus one and one, right? So, a root in the open interval minus one, one? No, no, careful! One over x is never zero. There is no root at all. So what went wrong? Continuity, yes! The theorem needs f continuous on the whole closed interval. But one over x is not continuous on minus one, one. It is not even defined at zero, and zero sits right inside this interval. So the theorem does not apply. Clear? So, the rule: always check continuity on the whole closed interval first, and only then look at the signs. Make it a habit. Okay?

4:14Example 1: show cos x = x has a root in (0, pi/2)

Now, remember the equation from the start? Cos x equals x. Let's finish it. Show that cos x equals x has a solution in the open interval zero to pi over two, where pi is this Greek letter. Pause the video, and try it first. Done? Okay, let's check. Our theorem talks about one function taking the value zero. But this equation has two sides. So which function should we test? Move everything to one side! Let g of x equal cos x minus x. Every solution of our equation is a root of g, right? And g is a difference of two continuous functions, so it is continuous on the closed interval zero, pi over two. Condition checked first! Now the left end. g of zero is cos zero minus zero, that is one, greater than zero. Positive. And the right end. Cos of pi over two is zero, so g of pi over two is zero minus pi over two, that is minus pi over two, less than zero. Negative! With me? So g changes sign. By the intermediate value theorem, g of c equals zero, that is, cos c equals c, for some c in the open interval zero, pi over two. See? No simple algebra solves it, but we proved a solution exists. Just as promised at the start!

5:43Example 2: continuous f from [0,1] to [0,1] has f(c) = c

Now a harder one. f goes from the closed interval zero, one, to the closed interval zero, one. The arrow means f takes inputs from the first and gives outputs in the second. And f is continuous. Show that f of c equals c, for some c. Pause the video and try it. Hint: use the same trick. Done? Okay, let's check. We want f of c equals c. So, move everything to one side: let g of x equal f of x minus x. f is continuous and x is continuous, so g is continuous on the closed interval zero, one. Okay? Now the left end. g of zero is just f of zero. But we have no formula for f! So what do we know about f of zero? Very good! Every output of f lies in the closed interval zero, one. So f of zero is at least zero. So g of zero is greater than or equal to zero; that is this sign, greater than or equal to. Right? Same way at the right end. f of one is at most one, so g of one, f of one minus one, is less than or equal to zero. Got it? So g goes from at least zero to at most zero, and zero lies in between. By the IVT, g of c equals zero, that is, f of c equals c. With me? Now, where is this c? Strictly between zero and one, in the open interval, right? Oh no, careful! Not necessarily. Our signs were greater than or equal to, and less than or equal to, not strict. So g of zero or g of one may itself be zero, and then c is an end point. That is why the answer says c belongs to the closed interval zero, one. Clear?

7:44Quick revision: IVT, sign change, check continuity first

So, quick revision. If f is continuous on the closed interval a, b, it takes every value between f of a and f of b. If such an f has opposite signs at the ends, there is a root in the open interval a, b. For an equation with two sides, move everything to one side, then check the signs. And continuity first, always: one over x on minus one, one changes sign, but has no root. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!