Current time:0:00Total duration:9:29
0 energy points

IIT JEE function maxima

Video transcript
"Let f, g, and h be real-valued functions defined on the interval, 0 to 1." And it includes the boundaries 0 and 1. That's what the brackets tell us. It wouldn't include the boundaries if we have parentheses. And then they define the functions. f of x is equal to e to the x squared plus e to the negative x squared. g of x looks very similar to f of x, except we have this x out here in front of the first term. And then we have h of x, which looks very similar to g of x and f of x, except it's got an x squared term right over there. And then they tell us, "If a, b, and c denote, respectively, the absolute maximum of f, g, and h on the interval, then?" So a is the maximum of f, b is the maximum of g, and then c is the maximum of h. And then they have a bunch of choices that compare the maximum values. Now, the first thing that may or may not jump out at you, just looking at these equations over here-- now, we're over the interval, 0 to 1. And let's just look at the functions themselves. Let's just look at the functions themselves. The difference is this one, you could view, has a 1 coefficient out front. This has an x, and this has an x squared. So over that interval, how are these functions going to relate? Over that interval. We know that over the interval, from 0 to 1, we know that x squared is going to be less than or equal to x, which is going to be less than or equal to 1. They all equal each other at the right interval, at actually being equal to 1. So this is going to be less than this, which is going to be less than that. And these are all positive values. And the rest of the functions are the same. This is the same as that, that's the same as that. And that is the same as that. So over the interval, we know that h of x is going to be less than or equal to g of x, which is going to be less than or equal to f of x. That just comes straight out of, this coefficient is going to be less than or equal to this coefficient, which is going to be less than or equal to 1, over the interval. Now with that said, let's actually try to figure out the maximum values. So I'll start with f. So f of x-- let me just tried it the boundaries. f of 0 is equal to e to the 0 squared plus e to the negative 0 squared. That's just e to the 0 plus e to the 0, which is 1 plus 1, which is 2. And then f of 1 is e to the 1, plus e to the negative 1. This is e to the 1, or just e, I should say, plus e to the negative 1. Now this is clearly larger than that. e is 2.71 and keeps going. e to the negative 1, that's still going to be a positive number, obviously. So this, between the two endpoints, this seems to be a candidate for the maximum point. We don't know that's the maximum point, because the function might have done something funky in between. So let me draw an axes. I can draw a straighter line than that. So we know at 0, we took on the value 2, and at 1, we took on the value e plus 1 over e, or e to the negative 1. So that's that value right over there. Now, we don't know what the function did in between. This would be the maximum point-- this is the maximum point, if the function did something like this. If it's constantly increasing the entire time, this would be the maximum point. It would not be the maximum point if the function looks something like that. And the way we can tell what the function is doing is we can take the derivative of f, which tells us the slope at any point over this interval. And if the derivative is constantly positive, that means we always have an upward slope. If the derivative is not always positive, then we have something like this and then we could actually look for that actual maximum point. So let's calculate the derivative. So f prime of x is equal to-- let's see the derivative of e to the x squared is 2x e to the x squared, just using the chain rule. Derivative of e to the negative x squared is negative 2x e to the negative x squared. And let's see, we could factor out-- this is equal to 2x e to the negative x squared times-- so if you divide this by e to the negative x squared, that's like multiplying it times e to the x squared. So this becomes e to the 2x squared minus 1. Now, our question is-- this is the slope over the interval-- what is this doing at any point? Well, this value right here, this right here is always going to be greater than or equal to 0. It'll be 0 at the left-hand interval. Just remember, this is the slope. The slope is 0 at the left-hand interval, which tells us it's either a maximum or minimum point. But then, it's going to be positive after that. Because this is going to be a positive value, and that this is going to be a positive value. Remember negative exponent does not mean a negative number. It means a fractional value. It's going be 1 over e to the x squared. So this is always going to be greater than or equal to 0. And so what we need to figure out, is this always greater than or equal to 0? So we want to figure out, is e to the 2 x squared minus 1 greater than or equal to 0 over the interval? Which is the same thing as asking, is e to the 2x squared greater than or equal to 1 over the interval? I just added 1 to both sides of this inequality. And that's the same thing as asking, if you take the natural log of both sides, the natural log of this is to 2x squared. The natural log of 1 is 0. So is 2x squared greater than or equal to 0 over the interval? Well, yeah, sure. It's going to be equal to 0 at x is equal to 0. But then after every other value, as it approaches 1, it'll be positive. So this is true. This is true. So this thing is always going to be greater than 0 over the interval, which tells us that the two parts of the derivative are always going to be greater than or equal to 0. So the derivative itself, the slope of the function, is always greater than 0. So always greater than 0-- over the interval I should say-- always greater than or equal to 0 over the interval from 0 to 1. Which tells us that the slope is never negative. And actually, we know that the slope is 0 at 0. So that tells us that the graph looks something like this, which tells us that the right endpoint-- that f of 1, when x equals 1-- f of 1 is the maximum point. So this, indeed, is equal to a. So we can actually write this here. f of x is less than or equal to a. Now, let's think about h of x and g of x. So f of 1 was its maximum point. But g of 1-- if you take g of 1 or h of 1, it'll evaluate to f of 1, because if you put a 1 here-- let's do that. Let me just do it to make it clear. So we already evaluated f of 1 right over there. g of 1 is equal to-- let me make sure we can see the-- so the g of 1 is equal to 1. 1 times e to the 1 plus e to the negative 1. That's g of 1, which is the same thing as f of 1. So this is also equal to a. And h of 1 is the same thing. So 1 times-- it's 1 squared, which is 1, times e to the 1 plus e to the negative 1, which is also equal to a. So what do we have, a situation here? We know that h of x over the interval is less than or equal to g of x, is less than or equal to f of x, which is less than or equal to a. And actually, the only point where they all equal each other is at x is equal to 1. If this right here is f of x, g of x is going to look something like this. And h of x is going to look something like this. h of x might look something like that. And so they all finally equal each other. The only place they equal each other is at a, which is the maximum of f. So that also has to be the maximum of g and h, because they can both get to that value. And they can't get to any value higher than that. That's an upper bound on them. So this is also equal to b-- remember b is just the maximum of g. And it's also equal to c, which is the maximum of h. So a, b, and c are all equal to each other, or we can say choice D.