Create a decomposition of functions

**mysisi** · October 8th 2012, 03:36 PM

Decomposing a function as a composition
Find f(x) and g(x) such that h(x)=(fog)(x)

h(x)=x²-3x/square root of x+5

I tried to do it but I do not think is right. this is the answer I get: f(x)= x²-3x g(x)= 1/square root of x+5

**johnsomeone** · October 8th 2012, 04:26 PM

Your answer was a quotient, not a composition. You found f and g so that h(x) = f(x)/g(x).

There are an unlimited number of choices given the way the problem is stated.
For instance, there are always these trivial choices:
Choice #1) f(x) = h(x) and g(x) = x
Choice #2) f(x) = x and g(x) = h(x)
Maybe more interesting is letting g(x) = x+5.

Then $h(x) = \frac{x^2-3x}{\sqrt{x+5}}$ and $x = g(x)-5$ , so

$h(x) = \frac{(g(x)-5)^2-3(g(x)-5)}{\sqrt{(g(x)-5)+5}}$

$= \frac{(g(x)^2 - 10g(x) + 25) + (-3g(x)+15))}{\sqrt{g(x)}}$

$= \frac{g(x)^2 - 13g(x) + 40}{\sqrt{g(x)}}$

$= f(g(x))$ , where $f(t) = \frac{t^2 - 13t + 40}{\sqrt{t}}$

Choice #3) $f(t) = \frac{t^2 - 13t + 40}{\sqrt{t}}, g(x) = x+5$

Now try $g(x) = \sqrt{x+5}$ . Note that $g(x) \ge 0$ (actually, it can't be allowed to be 0). Have $x = g(x)^2 - 5$ .

Thus $h(x) = \frac{x^2-3x}{\sqrt{x+5}}$ and $x = g(x)^2-5$ , so

Thus $h(x) = \frac{(g(x)^2-5)^2-3(g(x)^2-5)}{\sqrt{(g(x)^2-5)+5}}$

$= \frac{(g(x)^4-10g(x)^2+25) +(-3g(x)^2 + 15)}{\sqrt{g(x)^2}}$

$= \frac{g(x)^4 -13 g(x)^2 + 40}{|g(x)|}$ (now use $g(x) \ge 0$ to know that $|g(x)| = g(x)$ )

$= \frac{g(x)^4 -13 g(x)^2 + 40}{g(x)}$

$= g(x)^3 - 13g(x) + \frac{40}{g(x)}$

$= f(g(x))$ , where $f(t) = t^3 - 13t + \frac{40}{t}$

Choice #4) $f(t) = t^3 - 13t + \frac{40}{t}, g(x) = \sqrt{x+5}$

**mysisi** · October 9th 2012, 06:58 AM

Hi! Thanks for answering my question. However my professor said that is not the way he wants it. He said that the answer g(x) should be g(x)=1/square root of x+5 and I only need to find f(x). I guess he only wants a quotient? Mine is wrong.Can you help me please?? So when I find (fog) I will end up with: x^2-3x/ square root of x+5 back again. Thanks!!

**johnsomeone** · October 9th 2012, 07:04 AM

Review what I did for Choice #3 and Choice #4. In each case, I began with choice of g(x) (one time was x+5, the other it was sqrt(x+5)). Then I did some algebra and eventually found f(t) such that h(x) = f(g(x)). So if you repeat that process, you'll solve the problem. Repeat the process I did, only now your choice of g(x) will be:

$g(x) = \frac{1}{\sqrt{x+5}}$ .

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