Question

determine if h(x)=5/x + 2 is an inverse of
f(x) = 5/(x-2)

Answer 1

You can make this determination by re-writing f(x) as:

f(y) = x = 5/(y - 2)

and solve for "y" and see if you get the right side of h(x)

Answer 2

i got 5x+2 so im assuming the answer is no?

Answer 3

if its the inverse then h(f(x)) = x = f(h(x))

Answer 4

what is h(f(x))?

Answer 5

f(y) = x = 5/(y - 2)

y - 2 = 5/x

y = 5/x + 2

And that is the right side of h(x), so it is an inverse.

Answer 6

Original function:

Answer 7

inverse:

Answer 8

i dont see how you got that last conclusion...
how did you get y-2 over on the other side without multiplying with x?

Answer 9

Set the reciprocals equal to each other as a first step. It's faster and not really a short-cut.

Answer 10

1/x = (y-2)/5

Answer 11

Then, multiply both sides by 5:

5/x = y - 2 The last step, just add 2 to both sides

Answer 12

y=5/(x-2)
y(x-2)=5
yx-2y=5
yx=5+2y

x=(5+2y)/y
thus it is not the inverse...

Answer 13

@zzr0ck3r you have to switch the "x" and "y" first. Do that, THEN solve for "x" and you'll see that they are inverses.

Answer 14

they are not
(5+2x)/x is not equal to 5/(x+2)

Answer 15

They are not supposed to be.

Answer 16

I thought you were saying the two function in the question asked are inverses of each other, and they are not.

Answer 17

I just proved that they are inverses. To find out if functions are inverses, you take the original function, switch the variables and solve. You are forgetting to switch the variables first.

Answer 18

Look at the 2 functions, defined for x >= 0

y = x^2 and y = +sqrt(x)

You will see that they are not equal and are inverses

Answer 19

I thought 5/x + 2 = \[\frac{5}{x+2}\].... sorry

Answer 20

sorry @tcarroll010

Answer 21

np! You were eventually able to see it @zzr0ck3r

Answer 22

I thought you guys were crazy for a min:)

Answer 23

What do you mean? I'm crazy a lot! lol

Answer 24

I wish I could medal you twice:)

Answer 25

Another way to see the inverse relationship:

5 / [(5/x + 2) - 2] = x

and

5/[5/(x - 2)] + 2 = x

Answer 26

right that was my iff statement about f(h(x)) = x = h(f(x))

Answer 27

yes, I knew you knew about inverses and substituting back. You're way too good not to, so I got a little surprised when you didn't see it right away. I figured you misread the function and that was all.

Answer 28

I feel lame:)

Answer 29

No way! I wish I made as few mistakes as a lot of the others. No lameness! Only good discussions!

So, anyway, @nechamoosh are you all set yet?

Answer 30

oh. yeah sorry! thankss again for the help!!!

Answer 31

ha, both of you!

Answer 32

Yes, zzrock3r ' s contribution was definitely worthwhile and had a lot of good theory. His help was very valuable. And I'm glad we all learned something here. I sure did!

Answer 33

But now, it's Miller time. So, I'll stick around just briefly if you have any last minute questions, @nechamoosh So, fire away if anything is unclear. Otherwise, nice working with you!

Answer 34

Leaving on a bit of humor, this problem I was "dude". Last problem, I was a lifeguard. And here I thought I was just Sheldon!

Answer 35

um well there is one last thing i dont get.
if A = {l, m, n, o} and B = {10, 20, 30, 40}. is the set of ordered pairs {(l,30) (m,20) (n,10)} a function from set A to set B? why or why not?

Answer 36

i am completely lost on that one!

Answer 37

Yes, it's a function because each of the values in "A" (your independent variable set) takes on at most only one value. If, for instance, you had 2 different values to go with "m", like another point being (m, 70), then you wouldn't have a function. It would then be only a relation.

Answer 38

ohhhhh, okay i get it! thanks so much!!

Answer 39

Consider the function y = 2x. That's a function because for each "x", there is only one "y". The inverse is also a function, y = x/2 .Example of a function where the inverse is only a relation:

y = x^2

Answer 40

uw! So, if you have a new question, just start it in a new post and someone will come along. If you have any question with what we've covered, then it's ok to ask it here.

Answer 41

nah im all good! but thanks