Thread: 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)
>??????

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)
>??????

,



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





you should be able to see that

Originally Posted by skeeter

,



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





you should be able to see that

so it is right that i say

Originally Posted by ansonbound

so it is right that i say

not for all x ... only for x = 1

the last question..

swap variables ...



solve for y to finish

Originally Posted by ansonbound

yes! i did it by using

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?