1. ## Composite functions

Am I working this problem properly?
If $f(x)=3x+1$, and $g(x)=(x^2+5x)^{-\frac{1}{2}}$, find $g(f(x))$.

Working:
If $f(x)=3x+1$, and $g(x)=(x^2+5x)^{-\frac{1}{2}}$ then, $g(f(x))$= $((3x+1)^2+5(3x+1))^{-\frac{1}{2}}$.

Am I correct, and is that the final answer?

2. Originally Posted by koumori
Am I correct,
Yep

Originally Posted by koumori
and is that the final answer?
Expand inside the main bracket and group like terms to simplify

3. Originally Posted by pickslides
Yep

Expand inside the main bracket and group like terms to simplify

Ok...so the final answer looks like this:
$((9x^2+1)+(15x+5))^{-\frac{1}{2}}\rightarrow \frac{1}{\sqrt{(9x^2+1)+(15x+5)}}$?

4. Actually, what you should have found was that the expression simplifies to

$((9x^2 + 6x + 1) + (15x + 5))^{-\frac{1}{2}}$,

and that can be further simplified to

$(9x^2 + 21x + 6)^{-\frac{1}{2}}$.

You can also take out a factor of 3:

$(3(3x^2 + 7x + 2))^{-\frac{1}{2}}$

5. Originally Posted by koumori
Ok...so the final answer looks like this:
$((9x^2+1)+(15x+5))^{-\frac{1}{2}}\rightarrow \frac{1}{\sqrt{(9x^2+1)+(15x+5)}}$?
No, $(3x+1)^2 \neq 9x^2+1$

6. Originally Posted by icemanfan
Actually, what you should have found was that the expression simplifies to

$((9x^2 + 6x + 1) + (15x + 5))^{-\frac{1}{2}}$,

and that can be further simplified to

$(9x^2 + 21x + 6)^{-\frac{1}{2}}$.

You can also take out a factor of 3:

$(3(3x^2 + 7x + 2))^{-\frac{1}{2}}$
Originally Posted by pickslides
No, $(3x+1)^2 \neq 9x^2+1$
I see! Because I should use F.O.I.L on $(3x+1)^2$ making it
$((9x^2 + 6x + 1) + (15x + 5))^{-\frac{1}{2}}$. Then I have to do simple addition of like terms to net a result of $(9x^2 + 21x + 6)^{-\frac{1}{2}}$. Then do as you said and factor everything by 3.
Finally, expand the exponent so $(9x^2 + 21x + 6)^{-\frac{1}{2}}$ becomes: $\frac{1}{\sqrt{(9x^2 + 21x + 6)}}$

Funny how I lost my head at this once the big words came out.......

Thank you for your help guys!