Composition of fraction, please help

[lolilovepie]

can somebody please help me solve this problem. I tried looking in the book and find any problem similar to it, but cannot. Can you please help me solve this one problem. Thanks

the problem is:

Find and simplify the composition of f and g if:

f(x)=-2x^2+3, g(x)= (sq rt (2x-3))/ (2)
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So i tried solving it and i did (fog)(x)

and got 2(sqrt(2x-3))/(2))^2+3

and then i got stuck.. not sure if i'm doing it right either ..

 

[lolilovepie]

\(\displaystyle f(x)=2x^2+3\quad g(x)=\frac{\sqrt{2x-3}}{2}\)

you had the right idea

\(\displaystyle f\left(g(x)\right)=2(g(x))^2+3=2(\frac{\sqrt{2x-3}}{2})^2+3\)

do you know how to find \(\displaystyle (\frac{\sqrt{2x-3}}{2})^2\) ?

would \(\displaystyle (\frac{\sqrt{2x-3}}{2})^2\) become 2x-3/4?

 

[lolilovepie]

have to be careful with the ()'s but yes

\(\displaystyle (\frac{\sqrt{2x-3}}{2})^2=\frac{2x-3}{4}\)

we'd write that as (2x-3)/4 if we didn't want to be fancy

can you finish it from here?

I think so, so i did -2((2x-3)/(4))+3

and then got ((-4x+6)/(4))+3 (i cancelled the 4 out)

and then got 9-x as the answer

is that correct?

do I have to find the domain of some sort in the equation?

 

[lolilovepie]

why -2? Look at f(x) again.

oh the original problem had a negative 2

f(x)= - 2x^2+3, g(x)= (sqrt 2x-3)/2

 

[lolilovepie]

oh so it did... my eyes aren't what they used to be.

ok 9-x isn't quite right. Check your algebra.

ok so what i did was :

=-2((2x-3)/4))+3

=> ((-4x+6)/(4)) +3

=> (-4x+6+12)/4

=> -x+9

i'm not sure if i'm doing it correctly. I keep ending up with -x+9

 

[lolilovepie]

what's 18/4 ?

9/2,

where did the 18/4 come from?

 

[lolilovepie]

\(\displaystyle \frac{-4x+12+6}{4}=\frac{-4x+18}{4}=\frac{-4x}{4}+\frac{18}{4}=\frac{9}{2}-x\)

ohh okay, thanks a lot