Rolle's Theorem and the Mean Value Theorem

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Take f of x equals modulus x, on the interval from minus one to one. At both ends the value is one. The graph comes down to zero, and goes back up to the same height. So somewhere in between, the slope should be zero, right? But the slope of modulus x is minus one on the left side, and plus one on the right side. Never zero! So where did the flat point go? Hold that thought. Rolle's theorem and the mean value theorem answer it, and both come up often in JEE problems. So, let's begin.

0:40Rolle's theorem

So, Rolle's theorem. Three conditions, and each one matters. First, f is continuous on the closed interval a comma b, and second, differentiable on the open interval a comma b. Square brackets mean closed: both end points are included. Round brackets mean open: the end points are left out. So at the two ends we need only continuity. Okay? Third, f of a equals f of b. The function has the same value at both ends. Clear? Now think. The graph starts at some height, moves smoothly, and comes back to the same height. Somewhere in between, what must its slope be? Zero. Yes, exactly! f prime of c equals zero, for some c that belongs to the open interval a comma b. The little dash is read prime, and f prime is the derivative. The symbol like a rounded e means belongs to. So somewhere strictly between a and b, the curve has a horizontal tangent. Got it?

1:42The trap: check the conditions first

Now, back to our puzzle: f of x equals modulus x, on the closed interval minus one to one. The two vertical bars mean modulus. f of minus one is one, and f of one is also one. So the third condition holds. Okay? So, which condition fails? Go through all three again. Differentiability. Correct! Modulus x is continuous on minus one to one, no problem there. But at x equals zero, the left slope is minus one and the right slope is plus one. They do not agree, so modulus x is not differentiable at zero, and zero lies inside the open interval. Right? So wherever the derivative exists, f prime of c is plus or minus one, that is the plus minus sign you see, and never zero. Mystery solved: the theorem is not wrong, it simply does not apply. This is a classic JEE trap. Before using Rolle, check all three conditions. With me? And remember: differentiable implies continuous, but not the reverse. The double arrow means implies. Modulus x is the perfect example: continuous, but not differentiable at zero. Clear?

2:56Lagrange's mean value theorem

Now, the bigger theorem: Lagrange's mean value theorem, or LMVT for short. Only two conditions: f is continuous on the closed interval a comma b, and differentiable on the open interval a comma b. This time, f of a need not equal f of b. Right? Look at the figure. The orange line joins the two end points of the curve, at x equals a and at x equals b. We call it the chord. What is its slope? Rise over run. Correct! The slope of the chord is f of b minus f of a, over b minus a. Got it? And LMVT says: for some c in the open interval a comma b, f prime of c equals exactly this chord slope. With me? In the figure, look at the green dashed line. It is the tangent at the point c, and it is parallel to the chord. That is the whole theorem in one picture. Clear? Now a quick question. What happens if f of a equals f of b? The chord becomes flat, so its slope is zero. Then f prime of c equals zero, and that is exactly Rolle's theorem! So Rolle is a special case of LMVT, right?

4:09Two consequences

Now two quick consequences. First, if a and b are two roots of f, then f of a and f of b are both zero. Equal values! So, when f is continuous and differentiable, Rolle gives a c between them where f prime of c is zero. Between two roots of f lies a root of f prime. Okay? Second, rewrite LMVT: f of b minus f of a equals f prime of c, times b minus a. With me? So if the modulus of f prime never goes above capital M, the modulus of f of b minus f of a is at most capital M times the modulus of b minus a. This sign means less than or equal to. So a bound on the slope gives a bound on the change. Got it?

4:53Example 1: verify Rolle's theorem for x2 - 5x + 6 on [2, 3]

Alright, let's try one. Verify Rolle's theorem for x squared minus five x plus six, on the closed interval two comma three. Pause the video and try it. Check the conditions first! Done? Okay, let's check. It is a polynomial, so it is continuous on the closed interval two to three, and differentiable on the open interval. Two conditions done. Right? Third condition. f of two is four minus ten plus six, which is zero. f of three is nine minus fifteen plus six, also zero. Equal! So Rolle's theorem applies. Okay? Now differentiate. f prime of x equals two x minus five. Simple! So, where is the slope zero? Solve two c minus five equals zero. c equals five over two. Very good! And five over two is two point five, which lies between two and three. So c is inside the open interval, and Rolle's theorem is verified. Perfect!

5:56Example 2: find c of LMVT for x(x - 1)(x - 2) on [0, 1/2]

Now a harder one. Find c of the mean value theorem for f of x equals x, times x minus one, times x minus two, on the closed interval zero to one half. Pause the video and try it. Careful at the end! Done? Okay, let's check. Again a polynomial: continuous and differentiable everywhere, so LMVT applies. Okay? First the end values. f of zero is zero. f of one half is one half, times minus one half, times minus three halves. The small dots mean times. Two minus signs make a plus, so it is three over eight. Right? Chord slope: three over eight minus zero, over one half minus zero. Three over eight, divided by one half, is three over four. Clear? Now expand f: x cubed minus three x squared plus two x. So f prime of c is three c squared minus six c plus two, and LMVT sets it equal to three over four. With me? Multiply by four: twelve c squared minus twenty four c plus five equals zero. The quadratic formula gives c equals one, plus or minus root twenty one over six. The wavy equals sign means approximately: about zero point two three six, or one point seven six four. Okay? So, two answers, right? No, no, careful! The interval is zero to one half. One point seven six four is far outside, so we reject it. Only c equals one minus root twenty one over six, about zero point two three six, lies inside. That is our answer. Always keep only the values of c inside the open interval. Got it?

7:45Quick revision: Rolle and LMVT conditions, valid c

So, quick revision. Rolle: continuous on the closed interval, differentiable on the open interval, and equal end values, then f prime of c is zero. LMVT: just continuous and differentiable, then f prime of c equals the slope of the chord. And always check every condition, and keep only the c inside the open interval. Now try the practice problems on this page. All the best!