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Doing a little practice exercise, and came across the following question:

There are three functions: $f(x)$, $g(x)$, and $h(x)$.

Find $f(x)$ and $g(x)$, such that $h(x) = f \circ g(x)$ and $g(x) = \sqrt{4} + 8$.

The value of $h(x)$ is defined as $h(x) = (\sqrt{4} + 8) ^ 4$.

I'm not really sure how to approach the problem. I know I'm going to have to solve this algebraically, but I just can't seem to wrap my mind around the manipulation of functions as opposed to variables and numbers.

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What do you mean by h(x) = f * g(x) ? And by 8. h(x) do you mean $8h(x)$? (sqrt(4) + 8) ^ 4=$(\sqrt{4}+8)^4=10000$ –  Américo Tavares Sep 11 '10 at 20:27
    
This question is an absolute mess. Do you mean to tell us that $g(x)$ is given as a constant function? And $h$ as well? Or was there supposed to be an $x$ in the formulas? Or was that supposed to be $h(4)$ and $g(4)$, or something like that? –  Arturo Magidin Sep 11 '10 at 20:29
    
I personally have no idea what h(x) - f * g(x) means, which is part of the problem. I have an idea that it may be equivalent to f(g(x)), but I really don't know. I'll clean up the question a bit. –  Meta Sep 11 '10 at 20:29
    
Edited in the meantime. It remains only $8. h(x)$: is it $8h(x)$? –  Américo Tavares Sep 11 '10 at 20:30
    
@Americo: It was meant to be a period (end of sentence). –  Arturo Magidin Sep 11 '10 at 20:46

1 Answer 1

up vote 1 down vote accepted

The question, as currently posted (even after I edited it) is a mess. Are $g(x)$ and $h(x)$ meant to be constant functions? Was that supposed to be $g(4) = \sqrt{4} + 8$ and $h(4)=(\sqrt{4}+8)^4$, or was it supposed to be $g(x)=\sqrt{x}+8$ and $h(x)=(\sqrt{x}+8)^4$? Something else?

In any case: a composition is a function that is obtained by first applying one function, and then taking the output of that function and plugging it into the next function.

Take the expression $(\sqrt{4}+8)^4$. Suppose you were trying to actually compute this number. How would you go about it? First, we would take the number $4$ and plug it into the square root to obtain $\sqrt{4}$. The we would take the ouput of that, and add $8$. Finally, we would take the result we got from doing all that, and raise it to the fourth power.

So you can think of this as the process of doing three things, one after another, where the input to each of the steps is the output of the previous step:

  1. Take the square root of what you have;
  2. Add $8$ to what you have;
  3. Raise what you have to the fourth power.

Now, you are told to try to express this as the result of doing two things in succession: the first one is $g(x)$, and the second one will be $f(x)$. And you are told that the first thing you are doing is $g(x)=\sqrt{4}+8$. So, what else do you need to do to the output of this process in order to get to the final result? You need to raise what you are given to the fourth power. So, whatever $f(x)$ is, it should take the output of $g(x)$ and raise it to the fourth power.

So, what is $f(x)$?

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I realize that there is information missing. The question is from an online website called MathXL, which is used by my university in first year calculus courses, and is rendered exactly as it was given to me. This question is one in a set of exercises given by the professor to help refresh our memory on basic concepts. Note that this is not homework; these questions are purely for practice and we, as such, receive no marks. –  Meta Sep 11 '10 at 20:41
    
My first comment is, then, that someone doesn't know how to use MathXL, or that (as is more likely), the publisher has no idea what it is doing. My second comment is: ask your professor. The first person to go to when you don't understand material assigned by a professor is that professor. (S)He'll know exactly what you are supposed to know and not know, and be in a position to look at the question itself directly, which we cannot. –  Arturo Magidin Sep 11 '10 at 20:45
    
Either way, thank you for the help. It makes much more sense now, and was trivially easy to solve. –  Meta Sep 11 '10 at 21:13
    
@Meta: In that case, would you accept Arturo's answer (or your own, if it ended up being significantly different)? –  Larry Wang Sep 13 '10 at 10:05
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@Meta: "accept" in the sense of marking it as the "accepted answer". See the FAQ, math.stackexchange.com/faq, at the end of the "How do I ask a question here?" section. –  Arturo Magidin Sep 13 '10 at 14:13

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