# Thread: f * g * h

1. ## f * g * h

Hello,

Can you please verify that I did the correct operation?
$\displaystyle f(x) = 2x-1$
$\displaystyle g(x) = x^2$
$\displaystyle h(x) = 1 - x$

$\displaystyle 2(1-x)^2 -1$
$\displaystyle 2(1-x^2)-1$
$\displaystyle 2-2x^2-1$
$\displaystyle -2x^2-1$

Thanks!

2. Originally Posted by l flipboi l
Hello,

Can you please verify that I did the correct operation?
$\displaystyle f(x) = 2x-1$
$\displaystyle g(x) = x^2$
$\displaystyle h(x) = 1 - x$

$\displaystyle 2(1-x)^2 -1$
$\displaystyle 2(1-x^2)-1$
$\displaystyle 2-2x^2-1$
$\displaystyle -2x^2-1$

Thanks!
$\displaystyle 2x^2-4x+1$

3. Dear l flipboi l,

I think intead of $\displaystyle {2(1-x^2)-1}\mbox{ it should be }{2(1-2x+x^2)-1}$

4. Originally Posted by Sudharaka
Dear l flipboi l,

I think intead of $\displaystyle {2(1-x^2)-1}\mbox{ it should be }{2(1-2x+x^2)-1}$
Sudharaka,

actually, you're right...you just expanded (1-x^2).

5. $\displaystyle f(g[h(x)])=2(g[h(x)])-1=2([h(x)]^2)-1$

$\displaystyle =2(1-x)^2-1=2(1-2x+x^2)-1=2x^2-4x+1$

6. Dear l flipboi l,

No I did'nt expand $\displaystyle 2(1-x^2)-1,\mbox{ I expanded }{2(1-x)^2-1}$