[MOVED] Pre-Calculus: Find [f*g](x) and [g*f](x) for....

[ricjen]

Find [f*g](x) and [g*f](x) for each g(x) and g(x)
f(x) = xsquared - 1
g(x) = 5xsquared

This is what I've done so far:

[f*g](x) = fg(x))          [g*f](x) = g[f(x)]
         = (5x^2)^2 - 1             = 5(x^2 - 1)^2
         = 5x^4 - 1                 = (5x^2 - 1)^2
                                    = 5x^4 + 1

The teacher is telling me I need to simplify this.
Please help

 

[M98Ranger]

When doing these sorts of problems ....or any problem for that matter, but especially composite functions you must use correct order of operations.

\(\displaystyle fog(x) = (5x^2 )^2 - 1 = 25x^4 - 1\) You were on the right track, just pay attention to where the parenthesis need to go. Also there are a couple algebraic rules you broke when manipulating the function. Keep putting forth effort and algebra like this will be easy before you know it.

 

[steve_b]

[f*g](x) = fg(x))
= (5x^2)^2 - 1
= 5x^4 - 1 <== NO

(5x^2)^2 - 1 = 25x^4 - 1

This is the difference of two squares, so you can factor:

= (5x^2 + 1)(5x^2 - 1)

---------------

[g*f](x) = g[f(x)]
= 5(x^2 - 1)^2
= (5x^2 - 1)^2 <== NO

5(x^2 - 1)^2 = 5(x^4 - 2x^2 + 1)

Steve

 

[ricjen]

When doing these sorts of problems ....or any problem for that matter, but especially composite functions you must use correct order of operations.

\(\displaystyle fog(x) = (5x^2 )^2 - 1 = 25x^4 - 1\) You were on the right track, just pay attention to where the parenthesis need to go. Also there are a couple algebraic rules you broke when manipulating the function. Keep putting forth effort and algebra like this will be easy before you know it.

Thanks

 

[ricjen]

[f*g](x) = fg(x))
= (5x^2)^2 - 1
= 5x^4 - 1 <== NO

(5x^2)^2 - 1 = 25x^4 - 1

This is the difference of two squares, so you can factor:

= (5x^2 + 1)(5x^2 - 1)

---------------

[g*f](x) = g[f(x)]
= 5(x^2 - 1)^2
= (5x^2 - 1)^2 <== NO

5(x^2 - 1)^2 = 5(x^4 - 2x^2 + 1)

Steve

Thanks