# Composition of Functions

• Jul 13th 2007, 12:28 PM
ceasar_19134
Composition of Functions
I need help understanding whats going on...

Let f(x)=(x^2)-1 and g(x)=3x.

Find f * g.

1.(f * g)(x)=f(g(x))
2.---------=f(3x)
3.---------=((3x)^2)-1
4.---------=(9x^2)-1

I primarily dont understand line 3. How can 3x((x^2)-1)=((3x)^2)-1 ?

Perhaps you have a better way of explaining.

Trying to teach myself Algebra 2.
• Jul 13th 2007, 12:47 PM
red_dog
\$\displaystyle f(g(x))=f(3x)=(3x)^2-1=9x^2-1\$
• Jul 13th 2007, 01:25 PM
ceasar_19134
I can see that. I just dont understand the third line of my original post.

How do they get to that conclusion?
• Jul 13th 2007, 01:43 PM
red_dog
\$\displaystyle f(3x)\$ means that in the function expression \$\displaystyle x\$ is replaced by \$\displaystyle 3x\$.
So \$\displaystyle f(3x)=(3x)^2-1\$, not \$\displaystyle 3x^2-1\$
• Jul 13th 2007, 02:09 PM
Krizalid
Quote:

Originally Posted by ceasar_19134
I need help understanding whats going on...

Let f(x)=(x^2)-1 and g(x)=3x.

Find f * g.

\$\displaystyle (f\circ{g})(x)=[g(x)]^2-1\$

But \$\displaystyle g(x)=3x\$, so \$\displaystyle (f\circ{g})(x)=(3x)^2-1={\color{blue}9x^2-1}~\blacksquare\$

Did you get it now?
• Jul 13th 2007, 03:03 PM
ceasar_19134
Okay I get line 3 now, but how did you get (9x)^2 out of (3x)^2?
• Jul 13th 2007, 10:50 PM
CaptainBlack
Quote:

Originally Posted by ceasar_19134
Okay I get line 3 now, but how did you get (9x)^2 out of (3x)^2?

Nobody has got \$\displaystyle (9x^2)\$ from \$\displaystyle (3x^2)\$ what they
have done is: \$\displaystyle (3x)^2=(3)^2(x)^2=9x^2\$

RonL