# Thread: Verify that x = 1 => fg(x) = gf(x)

1. ## Verify that x = 1 => fg(x) = gf(x)

a)The functions f and g are defined by f(x) = root x for x > 0; g(x) = x - 1 for all values of x.

i.)Write down expressions for fg(x) and gf(x).
> fg(x) = root x - 1
> gf(x) = root x - 1

ii.) Verify that x = 1 => fg(x) = gf(x)
>??????

2. Originally Posted by ansonbound
a)The functions f and g are defined by f(x) = root x for x > 0; g(x) = x - 1 for all values of x.

i.)Write down expressions for fg(x) and gf(x).
> fg(x) = root x - 1
> gf(x) = root x - 1

ii.) Verify that x = 1 => fg(x) = gf(x)
>??????
$\displaystyle f(x) = \sqrt{x}$ , $\displaystyle x > 0$

$\displaystyle g(x) = x - 1$

I assume you mean function composition by the notation "fg(x) and gf(x)"

$\displaystyle f[g(x)] = f(x-1) = \sqrt{x-1}$

$\displaystyle g[f(x)] = g(\sqrt{x}) = \sqrt{x} - 1$

you should be able to see that $\displaystyle f[g(1)] = g[f(1)] = 0$

3. Originally Posted by skeeter
$\displaystyle f(x) = \sqrt{x}$ , $\displaystyle x > 0$

$\displaystyle g(x) = x - 1$

I assume you mean function composition by the notation "fg(x) and gf(x)"

$\displaystyle f[g(x)] = f(x-1) = \sqrt{x-1}$

$\displaystyle g[f(x)] = g(\sqrt{x}) = \sqrt{x} - 1$

you should be able to see that $\displaystyle f[g(1)] = g[f(1)] = 0$
so it is right that i say $\displaystyle \sqrt{x-1} = \sqrt{x} - 1 = 0?$

4. Originally Posted by ansonbound
so it is right that i say $\displaystyle \sqrt{x-1} = \sqrt{x} - 1 = 0?$
not for all x ... only for x = 1

5. the last question..

6. $\displaystyle y = \sqrt{x-1}$

swap variables ...

$\displaystyle x = \sqrt{y-1}$

solve for y to finish

7. Originally Posted by ansonbound
yes! i did it by using
$\displaystyle b^2 - 4ac = 0 16 - 4*k*2 = 16 k = 2$

8. sketch the graph y = |2x-4|

sketch the graph y = x

how far up will you have to translate y = |2x-4| so that it touches y = x at only one point?