g(sqrt[x + 4])^2 = x + 4, x GTE 4: why?

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I was solving for g composed with f, and my book shows the results in the following way. I totally don't get this.

f(x) = sqrt[x + 4] and g(x) = x^2, x > 4.

I understand the functions find, but why the "x > 4"? I have no idea why it was simplified this way. Thanks for any clues.

gcomposedfrb5.jpg


- nicholas
 
I was solving for g composed with f, and my book shows the results in the following way. I totally don't get this.

f(x) = sqrt[x + 4] and g(x) = x^2, x > 4.

I understand the functions find, but why the "x > 4"? I have no idea why it was simplified this way. Thanks for any clues.
Click to expand...
Are you sure it isn't "-4"? The author is making it clear that one cannot take the sqrt of a negative number; i.e., the domain of x must be greater than or equal to -4.
 
Yeah I know right? I totally keep posting goofed up images but this time it was + 4. The original question asked the following:

given
fxtp0.jpg
and
gxcp7.jpg
, find
fcomposedgqb2.jpg


Thanks for replying.
 
unregistered said:
The original question asked the following: Given \(\displaystyle f(x)\,=\,\sqrt{x\,+\,4}\) and \(\displaystyle g(x)\,=\,x^2\), find \(\displaystyle f\,\circ\,g\).
Click to expand...
You say the books asked for f(g(x)), but then gives the answer for g(f(x)). From this, I would have to assume that the "answer" listed for this exercise is a typo.

Eliz.
 
Ah crap, I can't even get this straight, I'm sorry:

given
fxrj8.jpg
and
gxrz9.jpg
, find
gcomposedfkg4.jpg
.

The entire equation:

entiregk4.jpg
 
Since the book gives the working as:

. . . . .\(\displaystyle \L (g\,\circ\,f)(x)\,=\,g(f(x))\,=\,g(\sqrt{x\,+\,4})^2\)

...rather than the proper:

. . . . .\(\displaystyle \L (g\,\circ\,f)(x)\,=\,g(f(x))\,=\,g(\sqrt{x\,+\,4})\,=\,\left(\sqrt{x\,+\,4}\right)^2\)

....I still vote for a typo.

Eliz.
 
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