### Assignments:

Unfinished Assignment Study Questions for Lesson 9

### Lesson Objectives:

- Find the composition of two functions and its domain.
- Decompose a function as a composition of two functions.

A composite function is a function which depends on the output of another function for its input.

The composition of f and g, is defined as (f@g)(x), which is the same as f(g(x)), where x is the domain of g, and g(x) is the domain of f.

Given three functions, f(x) = 2x+1, g(x) = x^2-3x-8 and h(x) = x^3.

Find (f@g)(-1). f(g(-1)). We have to find what g(-1) is first. So plug -1 into g(x), and get -1^2-3(-1)-8. So we get 1+3-8, which is equal to -4. So now we need to find f(-4), by plugging -4 into f(x). So f(-4) is 2(-4)+1, which -8+1, which is -7. So (f@g)(-1) = -7

Next let's find (h@h)(x). This is the same as h(h(x)), so since h(x) = x^3, we have to plug x^3 into h(x). So h(x^3) is the same as (x^3)^3. We have to multiply 3*3 to get 9, so it's equal to x^9. (h@h)(x) = x^9.

Now find (f@g)(x) and (g@h)(x) and their domains. (f@g)(x) is the same as f(g(x)), where g(x) is x^2-3x-8. So now we have to plug all of that in over here, and get 2(x^2-3x-8)+1, which is 2x^2-6x-16+1, which is 2x^2-6x-15. Since this function does not have a radical or a fraction, that means that the domain is all real numbers, (-oo, oo).

Now, (g@h)(x) is the same as g(h(x)), where h(x) is x^3. So g(x^3), we have to plug in x^3 in for x over here. And we get (x^3)^2-3(x^3)-8 or x^6-3x^3-8. And since once again we don't have a radical or a fraction, the domain is all real numbers, (-oo, oo).

Find f(x) and g(x) such that h(x) = (f@g)(x). Answers may vary.

So h(x) = (5+2x)^7. We're trying to find (f@g)(x), which is the same as f(g(x)).

So g(x) = 5+2x and f(x) = x^7, then this will make h(x) = (f@g)(x).

Now if h(x) = 1/((x-4)^2), and we want this to equal (f@g)(x), which is the same as f(g(x)), then we have to let g(x) = x-4 and f(x) = 1/(x^2) for h(x) = (f@g)(x).