1. 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.

2. $\displaystyle f(g(x))=f(3x)=(3x)^2-1=9x^2-1$

3. I can see that. I just dont understand the third line of my original post.

How do they get to that conclusion?

4. $\displaystyle f(3x)$ means that in the function expression $\displaystyle x$ is replaced by $\displaystyle 3x$.
So $\displaystyle f(3x)=(3x)^2-1$, not $\displaystyle 3x^2-1$

5. 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?

6. Okay I get line 3 now, but how did you get (9x)^2 out of (3x)^2?

7. 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