Composition of Functions

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July 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.
July 13th 2007, 12:47 PM
red_dog

$f(g(x))=f(3x)=(3x)^2-1=9x^2-1$
July 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?
July 13th 2007, 01:43 PM
red_dog

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

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

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

Did you get it now?
July 13th 2007, 03:03 PM
ceasar_19134

Okay I get line 3 now, but how did you get (9x)^2 out of (3x)^2?
July 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 $(9x^2)$ from $(3x^2)$ what they
have done is: $(3x)^2=(3)^2(x)^2=9x^2$

RonL

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