Arithmetic Of Functions

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nolly

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Jul 17, 2007
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Hey guys this is my first time so please bear with me. To be honest im not even sure how i should start this problem, im really stuck lol.

Here is what they want me to do:

Find [h(x)] squared (over)
f(x)


The values for f(x) = x+2

The values for h(x) = 1 (over)
x-1

Im really sorry if i cause any confusion and Thank u so much for taking the time help people like me :D
 
\(\displaystyle \L f(x) = x + 2\quad \& \quad h(x) = \frac{1}{{x - 1}}\)

\(\displaystyle \L \frac{{h^2 (x)}}{{f(x)}} = \frac{{\left( {\frac{1}{{x - 1}}} \right)^2 }}{{x + 2}}\)
 
ok i follow u so far.

so now i come up with:

1
--------------------------
x^2 - 2x + 2
--------------------------
x+2



then:

(x-1)^2(x+2)
--------------------------
1


i think thats the solution??
 
\(\displaystyle \L f(x) = x + 2\quad \& \quad h(x) = \frac{1}{{x - 1}}\)


You are trying to find:

\(\displaystyle \L \frac{{h^2 (x)}}{{f(x)}} = \frac{{\left( {\frac{1}{{x - 1}}} \right)^2 }}{{x + 2}}\)

What you found is:

\(\displaystyle \L \ \frac{{x + 2}}{{\left( {\frac{1}{{x - 1}}} \right)^2 }}\)

See the difference!
 
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