>>Here we’re going to find

the composition of f with g and the composition of g

with f, and their domains, where f of x is, 2 to the

x and g of x is, x plus 4. So first of all, we’ll

look at f of g of x. Which is what f takes g of x to. Since g of x is, x plus 4 and f

uses its input as the exponent on 2, we get 2 to

the power, x plus 4. There is no restriction

when x goes to, x plus 4, or when we exponentiate.

Both, have domains which are all reals, and

so the domain of the composition of f with g is equal to

negative infinity, to infinity. Now as we look at g..

composition of g with f, of x. That is what

g takes f of x to. Where f of x is, 2 to the x. And

what does g to do to anything? g adds 4, and so

we’re going to have 2 to the x quantity, plus 4. This time x first of

all gets exponentiated, and.. I’m sorry.. I wrote

e instead of 2 here. And then we add 4, and so

there’s no restriction on either of these, the domain

of the composition of g with f is negative

infinity, to infinity. In our second problem, f

of x is log base 4 of x, and g of x is, x minus 4. So the composition of f with g,

of x, is what f takes g of x to.. what f takes, x minus 4 to and f takes the log base

4 of whatever it sees. So we get a log base

4 of, x minus 4. So x first of all

goes to, x minus 4, which allows any real number. But then to log base

4 of, x minus 4. And here we have a restriction.

x minus 4, must be greater than 0 because the log only

accepts positive values. So x minus 4, greater than

0 implies x must be greater than 4, and so the domain of the

composition of f with g is equal to the open interval 4 to

infinity, not including the 4. For the composition

of g with f of x, the result is what

g takes f of x to. And so this will be g of log

base 4 of x, g subtracts 4 from anything it sees. So it’s the log base 4 of

x quantity, subtract 4. This time x first of all is

taken to log base 4 of x, and then 4 is subtracted

from log base 4 of x. So the restrictions

occurs in the first step, where x must be positive. Because the domain of the

log base 4 is 0 to infinity, and so the domain of

the composition of g with f is the open

interval from 0 to infinity. There’s no further restriction because we can subtract

4 from any real number. And our last problem, we have f

of x equals log x.. common log x.. and g of x equals

x to the fifth.. it’s called a power

function.. x to the fifth, where the variable

is in the base. So composition of f

with g of x takes x to what f takes g of x to. And g of x is x to the fifth. f is the common log, so we end up with the log of

x to the fifth. So x goes to x to the fifth, which goes to the log

of x to the fifth. So x to x to the fifth, x

could be any real number, but in the second step x to the

fifth must be greater than 0. If x to the fifth

is greater than 0, we notice that a negative number

to the fifth would be negative, and so we must have

x greater than 0, because we’re not allowed to have a negative power

as an input to log. And so the domain of

the composition of f with g is the open

interval from 0 to infinity. In the other direction we have

the composition of g with f of x, which is what

g takes f of x to, and f of x is the

common log of x. And g raises whatever it

sees to the fifth power. So we’re going to have

the common log of x, parentheses to the fifth power. So x goes to log x, which goes to log x quantity

to the fifth power. Here the restriction occurs

right away, because the input to the log has to

be greater than 0. We can raise any number

to the fifth power, so there’s no further

restriction, and so the domain of the composition of g with f is the open

interval from 0 to infinity.

How about

f(n) = n log(n+2)

g(n) = n^(1.3)

I'm gonna kill this fucking bitch.