00:04
This problem looks at systems of equations and plugging in various values from one function to the other function.
00:16
And it also covers inverse functions.
00:19
So let's get started.
00:21
We're going to change up the example here.
00:23
So let's try this.
00:26
We're going to have our g is going to be x plus 2.
00:35
And let's say f, which is a second function, is going to be 5x plus 3.
00:55
Now, the first thing we're going to do is we're going to go ahead and find f of negative 1.
01:04
So all that means is we're actually just going to plug in negative 1 here.
01:08
So let's do this.
01:10
What is f of negative 1? so all that is saying is we're just going to plug in negative 1 for x in our function here.
01:23
So that tells us that this is going to be negative 1 times 5, which is negative 5 plus 3.
01:32
So that tells us this is negative 5 plus 3 tells us that our f of minus 1 value is actually minus 2.
01:45
There we go.
01:46
Now let's look at the second part of this question.
01:49
It says find g of f of 2.
01:54
In this case, g of f of negative 1, right? and all this is saying is we're going to take the value that we just found, and we're just going to plug it in right here.
02:16
So we've already done this intersection.
02:22
That is already done.
02:23
So what we can do is we can actually just go ahead and plug in this negative 2 here...