Composite function help ? - Thanks

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nukeman
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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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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.
 
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 :(
 
Yep, that's correct, except it's f(g(x)), not f(x).
 
Really? lol cool.

Ok, so then after that, what do I do? This is where I get very confused.
 
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]
 
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?
 
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:
I think I got it...

[-5, -1]

?
 
nukeman said:
I think I got it...

[-5, -1]

?

Yes, that's right.
 
Yep, I think that's right.