Decomposing a function as a composition

Find f(x) and g(x) such that h(x)=(fog)(x)

h(x)=x^{2}-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)= x^{2}-3x g(x)= 1/square root of x+5

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- Oct 8th 2012, 03:36 PMmysisiCreate a decomposition of functions
Decomposing a function as a composition

Find f(x) and g(x) such that h(x)=(fog)(x)

h(x)=x^{2}-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)= x^{2}-3x g(x)= 1/square root of x+5 - Oct 8th 2012, 04:26 PMjohnsomeoneRe: Create a decomposition of functions
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 $\displaystyle h(x) = \frac{x^2-3x}{\sqrt{x+5}}$ and $\displaystyle x = g(x)-5$, so

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

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

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

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

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

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

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

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

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

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

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

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

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

Choice #4) $\displaystyle f(t) = t^3 - 13t + \frac{40}{t}, g(x) = \sqrt{x+5}$ - Oct 9th 2012, 06:58 AMmysisiRe: Create a decomposition of functions
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!!

- Oct 9th 2012, 07:04 AMjohnsomeoneRe: Create a decomposition of functions
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:

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