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# 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.