Composite function help ? - Thanks

In summary, the homework statement is that the problem is that the student cannot seem to solve the first equation. They need help from the very beginning and the student provides a summary of the content.
  • #1
nukeman
655
0

Homework Statement



Here is the problem:

2wntl60.png



Homework Equations





The Attempt at a Solution



I need help RIGHT from step one.

Now, step one I would suppose I need to evaluate g(x) ? Which, would be x must be greater than or equal to -5. Correct?

Then what?
 
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  • #2
The first step is to write the function f(g(x)). What this means is that for the function f(x), you substitute x=g(x). Then, you need to find the domain of this f(g(x)) function.
 
  • #3
Ok, so first thing I would do is write in the g(x) into the f(x) ?

So, it turns into: f(x) is 3(2-(x + 5)^1/2)^(1/2)

? That can't be correct :(
 
  • #4
Yep, that's correct, except it's f(g(x)), not f(x).
 
  • #5
Really? lol cool.

Ok, so then after that, what do I do? This is where I get very confused.
 
  • #6
Neither square-root can have a negative number inside, so to find the domain, you need to find the intervals of x that make both the inner and outer square-roots positive.

In other words, x needs to satisfy the conditions:
[itex]\displaystyle x+5≥0[/itex]
and
[itex]\displaystyle 2-\sqrt{x+5}≥0[/itex]
 
  • #7
So this is all I do to find the formula, as asking in the question. f(x) is 3(2-(x + 5)^1/2)^(1/2)

x + 5 >= 0 would just be [0, infinity) right? is that what you mean?

Now sure how you got the 2nd one? what about the 3?
 
  • #8
nukeman said:
So this is all I do to find the formula, as asking in the question. f(x) is 3(2-(x + 5)^1/2)^(1/2)

x + 5 >= 0 would just be [0, infinity) right? is that what you mean?

Now sure how you got the 2nd one? what about the 3?

The 3 is on the outside of the 2nd square root. You're just trying to make the inside of the square root larger than 0, so the 3 can be disregarded when finding the domain.

In order to solve the inequalities, you treat it just like any other equation. To solve the first one, you subtract 5 from both sides to get:

[itex]\displaystyle x+5 ≥ 0[/itex]

[itex]\displaystyle x ≥ -5[/itex]

Hopefully that gives you an idea how to solve the second inequality. After you find the solution to both of those, your domain will be the x values that satisfy both inequalities.

nukeman said:
f(x) is 3(2-(x + 5)^1/2)^(1/2)
Also, remember that the composite function is not f(x), but f(g(x)). (also written as [itex](f \circ g)(x)[/itex])
 
Last edited:
  • #9
I think I got it...

[-5, -1]

?
 
  • #10
nukeman said:
I think I got it...

[-5, -1]

?

Yes, that's right.
 
  • #11
Yep, I think that's right.
 

1. What is a composite function?

A composite function is a combination of two or more functions where the output of one function is used as the input for another function. It is denoted by f(g(x)) and read as "f composed with g of x".

2. How do you find the domain of a composite function?

To find the domain of a composite function, you need to consider the domains of each individual function and determine where they overlap. The domain of the composite function will be the set of all inputs that are valid for both functions.

3. What is the difference between a composite function and a regular function?

A composite function is made up of multiple functions that are connected together, while a regular function is a single function with its own set of inputs and outputs. In a composite function, the output of one function becomes the input of another function.

4. Can a composite function have more than two functions?

Yes, a composite function can have any number of functions as long as the output of one function can be used as the input for the next function. However, it is more common to see composite functions with only two functions.

5. What is the purpose of using composite functions in mathematics?

Composite functions allow us to combine and manipulate multiple functions to create more complex relationships between variables. They are useful for solving problems in various fields such as physics, engineering, and economics.

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