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Here we're going to find some more composite functions.
And this time, one of our functions will be a square
root and the other will just be a linear function.
And we're going to have to be a little bit careful about
domain when we are dealing with something like a square
root function.
Or also if you had a function like a rational function where
you had a variable in a denominator, you'd also want
to be very careful about domains, or in other words,
the allowed input values.
So let's find f of g of x, g of f of x, and f of g of 2.
So for f of g of x, remember that that means f of g of x.
So we're going to use g as the input for f.
So we start with f.
f of x is equal to the square root of x.
And now if we're using g of x as the input, f of g of x,
that means replace x by the function g of x.
So I've still got the square root, but I'm going to replace
the x now with what g of x equals, which is x minus 1.
And that's it.
There's no simplifying to do.
So this is the same thing.
It's just a different notation for f of g of x.
So here's our composite function.
f of g of x is equal to square root of x minus 1.
Now, as far as the domain, there's a couple of things to
watch out for.
Because we made f of g of x, we want to make sure that we
locate or identify any unallowed input values of x
for the function g of x.
So if we look at that, g of x is x minus 1.
There's no unallowed values there.
We can use all real numbers.
So we're all set there.
No restrictions.
Then we also want to make sure that the value g of x is an
allowed input for the function f.
Now, f of x is root x.
So when we have g of x as the input, we have square
root of g of x.
So in other words, we want the g of x to be greater than or
equal to 0.
It can't be a negative number, because then we'd have a
square root of a negative.
Well, g of x is x minus 1.
And we need to be greater than or equal to 0.
So x must be greater than or equal to 1.
So that would be a restriction that you have on your domain
for this function.
In part B, we're finding g of f of x.
And remember that means g of f of x.
So we're using f as the input for g.
That means we start with g, g of x is equal to x minus 1.
We put f of x in as the input.
So everywhere you see an x, put f of x instead.
f of x is root x.
And then bring down the minus 1.
And that would be it.
There's no simplifying.
So that's just another way of writing g of f of x.
And there's our composite function, root x minus 1.
And again here, we want to think about what are the
allowed values for x.
Or think the flip side, are there any
unallowed values for x?
And because we're finding g of f of x, the first thing we
have to look at are there any unallowed values for the input
x to the function f?
Well, the function f of x is root x.
That means x must be greater than or equal to 0.
So there's one restriction that we have.
And then we're going to take the results of f of x and use
those as inputs for g.
So if g of x is x minus 1, what we want to look at is g
of f of x is, instead of x, we're putting in f of x.
So we have f of x minus 1.
So we need to think about are there any f of x values that
are not allowed?
And no, there aren't, because this isn't sitting in a square
root, it's not in a denominator,
so that's no problem.
We don't have any restrictions on the input for the g of x
function, or g of f of x.
But we still do have the restriction that x must be
greater than or equal to 0.
And the last part, we want to find f of g of 2.
And we can think of that also as f of g of 2.
So let's start with f of g of x, which we found in part A. f
of g of x is equal to square root of x minus 1.
And we'll replace x by 2.
f of g of 2 is equal to square root of, instead of x we have
2, minus 1.
So that's equal to square root of 1, or just 1.
And that would be f of g of 2.
And you can try it out, but you should get the exact same
answer, or you will get the same answer, if you first find
what's g of 2.
And once you find g of 2, take that result and use it as the
input for f, and you'll still get 1 as
your f of g of 2 value.