Absolute Extrema on a Closed Interval

Transcript

0:00Introduction

Hello everyone! Let us start with a small trek. You walk up a mountain trail, from the gate at the bottom to a viewpoint at the far end. Halfway along, you cross a small hilltop: the path climbs, goes flat for a moment, and then dips down again. Now, is that hilltop the highest point of your whole walk? Keep that question in mind. Today we find the greatest and the least value of a function on a closed interval, and that little hilltop hides the main trap. In JEE, questions on this come up often, and they are quick marks once you know the method. So, let's begin.

0:38Local peak or the highest point?

So, here is a graph from the last lesson: y equals x cubed minus three x, with its two turning points. The note at the top says f prime equals zero at both points. f prime is the derivative, the slope of the tangent, so the tangent is flat at both points, okay? At x equals minus one, the orange dot, we have a local maximum of height two. At x equals one, the green dot, a local minimum of height minus two. The dashed lines mark those two heights. Now tell me. Is two the greatest value this function ever takes? No, not at all! Look at the right side of the curve. It climbs past two at the right edge of the picture, and as x grows, it keeps rising without any limit. So a local maximum is the greatest value only near its own point, in a small neighbourhood. That is all the word local promises. Right? But a question usually gives us one fixed piece of the graph: a closed interval, from a to b, okay? The square brackets mean both ends are included. The absolute maximum is the greatest value of f on all of that interval. And the absolute minimum is the least value on all of that interval. Just like our trek: the highest and the lowest point of the entire walk, not of one small hill. Clear?

2:02Finding the absolute max and min

Now the main result. Suppose f is continuous on the closed interval from a to b, okay? Two conditions here. Continuous, so the graph has no breaks. And closed, so both ends are included. Then f is sure to attain an absolute maximum and an absolute minimum on that interval. This double arrow means implies. So the greatest and the least values really exist; our only job is to find them. With me? So, where can the greatest value sit? Think of our trek again. Yes, exactly! Only two kinds of places. At a critical point c inside the interval, or right at one of the two ends. These are our candidates. See the round brackets here? They mean the open interval from a to b, ends left out, because the ends are listed separately. Okay? And what is a critical point? From the last lesson: a point where f prime of c is zero, like a flat hilltop, or where f prime of c does not exist, like a sharp corner. Got it? Last step: compare. Write f of a, f of b, and f at every critical point. The largest value is the absolute maximum, the smallest is the absolute minimum. And here is a topper's trick. You do not need the second derivative test here at all. No need to decide which point is a local max or a local min. Just list the values and compare. Really simple!

3:28Example 1: x cubed - 3x on [0, 2], greatest value at an end

Now, Example one, and it uses the very graph on screen. Find the absolute extrema of x cubed minus three x on the closed interval from zero to two. Extrema simply means the maximum and the minimum together. Pause the video, and try it first. Done? Okay, let's check. f prime of x is three x squared minus three. Set it equal to zero: x squared is one, so x is plus or minus one. That is the plus minus sign you see on screen, okay? Now, do we keep both these critical points? No, only one! Minus one is not inside the interval from zero to two, so we drop it. On the graph, the orange local maximum at minus one lies outside our piece of the curve. Only x equals one lies inside. Clear? Now the values. f of zero is zero. f of one is one minus three, so minus two. And f of two is eight minus six, which is two. Three values: zero, minus two and two. Right? So, the maximum is two, at the end x equals two, and the minimum is minus two, at x equals one, the green dot. See? The greatest value came from an end, not from a flat top. Got it?

4:48Example 2: 2x cubed - 15x squared + 36x + 1 on [1, 5]

Now a harder one. Find the absolute extrema of two x cubed, minus fifteen x squared, plus thirty six x, plus one, on the closed interval from one to five. Pause the video and try it. Go slowly with the arithmetic! Done? Okay, let's check. f prime of x is six x squared, minus thirty x, plus thirty six. Take six common, and factorise: six times x minus two, times x minus three. Okay? So the critical points are x equals two and x equals three. Are they inside the interval from one to five? Yes, both of them. So we keep both. Right? Now the values. f of two: sixteen, minus sixty, plus seventy two, plus one. That gives twenty nine. f of three: fifty four, minus one hundred and thirty five, plus one hundred and eight, plus one. That gives twenty eight. With me? Many students stop right here and write: maximum twenty nine, minimum twenty eight. Is that the answer? No, no, careful! That is the trap. We forgot the two ends. f of one: two, minus fifteen, plus thirty six, plus one. That is twenty four. f of five: two hundred and fifty, minus three hundred and seventy five, plus one hundred and eighty, plus one. That is fifty six! So, compare all four values: twenty four, twenty nine, twenty eight and fifty six. The maximum is fifty six, at x equals five, and the minimum is twenty four, at x equals one. Both at the ends! Remember our trek? Twenty nine, at x equals two, is like that small hilltop. But the trail climbs to fifty six at the very end. That's it!

6:42Quick revision: closed interval method, never skip the ends

So, quick revision. Continuous on a closed interval from a to b: the absolute maximum and minimum exist. The candidates are the critical points inside, and both ends. The largest value is the maximum, the smallest is the minimum. And never, ever forget the ends! Now go and try the practice problems on this page. You will find them easy. All the best!