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

• June 27th 2010, 03:43 PM
ansonbound
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)
>??????
• June 27th 2010, 04:15 PM
skeeter
Quote:

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)
>??????

$f(x) = \sqrt{x}$ , $x > 0$

$g(x) = x - 1$

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

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

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

you should be able to see that $f[g(1)] = g[f(1)] = 0$
• June 27th 2010, 04:20 PM
ansonbound
Quote:

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

$g(x) = x - 1$

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

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

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

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

so it is right that i say $\sqrt{x-1} = \sqrt{x} - 1 = 0?$
• June 27th 2010, 04:30 PM
skeeter
Quote:

Originally Posted by ansonbound
so it is right that i say $\sqrt{x-1} = \sqrt{x} - 1 = 0?$

not for all x ... only for x = 1
• June 27th 2010, 04:58 PM
ansonbound
• June 27th 2010, 05:04 PM
skeeter
$y = \sqrt{x-1}$

swap variables ...

$x = \sqrt{y-1}$

solve for y to finish
• June 27th 2010, 05:53 PM
ansonbound
• June 27th 2010, 06:02 PM
ansonbound
Quote:
yes! i did it by using
$b^2 - 4ac = 0
16 - 4*k*2 = 16
k = 2$
• June 27th 2010, 06:05 PM
skeeter
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?