# F (f inverse (5)) = undefined?

1. Feb 5, 2005

### ktpr2

let f(x) = 3+x^2+tan(pi x/2), where -1 < x < 1

-> find f(f_inverse(5))

The teacher says f(f_inverse(5)) = 5

but I think f(f_inverse(5)) = undefined

I don't think the teacher considered the domain of f. You can't get 5 from -1 < x < 1, so who's wrong?

2. Feb 5, 2005

### Hurkyl

Staff Emeritus
What's the domain of f^-1?

3. Feb 5, 2005

### ktpr2

the domain of f^-1 is the range of f, which doesn't include 5 (when I graphed it)

Is there a way i can prove that f doesn't include 5 or can i just say something like 5 is not a member of the range of f?

4. Feb 5, 2005

### learningphysics

What is the domain and range of f?

What is the domain and range of f_inverse?

If f has an inverse and a is in the range of f... then f(f_inverse(a))=a by the very definition of the inverse.

Last edited: Feb 5, 2005
5. Feb 5, 2005

### ktpr2

That's my point; a = 5, and with a domain of -1<x<1 for f, a is NOT in the range of f, so f(f_inverse(a)) must be undefined, right?

6. Feb 5, 2005

### learningphysics

You made a mistake somewhere. 5 should be in the range of f.

7. Feb 5, 2005

### Hurkyl

Staff Emeritus
What's the range of tan (pi x/2) for -1 < x < 1?

8. Feb 5, 2005

### dextercioby

Yes,trust me,5 is definitely in the range of "f".I would go as far as telling you that the range of "f" is the entire R.

Can u see why??

Daniel.

9. Feb 5, 2005

### ktpr2

EDIT- remember the domain was restricted to (-1 < x < 1); if you stick the equation in a TI-8X you'll see that 5 is not in the range of f.

Okay I think can "prove" this. Be gentle cause I'm just taking calculus:

For all one to one functions we have,

a) f^-1(y) = x <-> f(x) = y;
and
b) f(x1) cannot equal f(x2), whenever x1 does not equal x2.

so,
f(x) = 5 when x = +/- 1.40056553...

But x is not a member of (-1 < x < 1)
(I can stop here)
and
f(-x) = f(x), so when f(x) = 5, f(x) is not a one to one function by (b). Thus, f(f^-1(5)) is undefined.

10. Feb 5, 2005

### Hurkyl

Staff Emeritus

11. Feb 5, 2005

### dextercioby

Take the calculator & compute the alue for x=0.999.

Daniel.

12. Feb 5, 2005

### ktpr2

I guess what I'm trying to say is that given the domain of -1 < x < 1, no one can give me a number that will make 3+x^2+tan(pi x/2) = 5. If I'm wrong, then please share that number.

I understand that tan has R for its range, but there are other components to the function, and with the restriction of the domain, you can't get the function to equal 5.

13. Feb 5, 2005

### Hurkyl

Staff Emeritus
So tell me, if the range of tan is all of R... how can adding a small value (how big can 3 + x^2 be?) keep f from reaching 5?

14. Feb 5, 2005

### dextercioby

Okay,Hurkyl,i'll leave it up to you...

Daniel.

15. Feb 5, 2005

### ktpr2

This is funny because I'm seeing both sides of the picture here. But...

While the range of tan may be R, the range of tan under a -1<x<1 is limited; it sure isn't R. If you look at this pretty picture here http://www.ping.be/~ping1339/tan.gif [Broken] you'll see that tan doesn't hit 5 until after 1 (or -1).

If I'm wrong then please give me a number that's w/in the domain that will make the function equal five. That's all I ask, a simple request.

Last edited by a moderator: May 1, 2017
16. Feb 5, 2005

### Hurkyl

Staff Emeritus
Yes, there is something funny going on...

You drew the graph of tan x, and saw that it reached 5 sometime after 1... but what about tan (pi x/2)?

17. Feb 5, 2005

### ktpr2

The number is 0.6421... I must have graphed it wrong. My apologies.

Wow I see what went wrong; I graphed the function in degree mode, not radians, which is why i got a "bad" graph

Last edited: Feb 5, 2005
18. Feb 5, 2005

### Hurkyl

Staff Emeritus
That's the answer I got. I hypothesized your calculator was in degree mode.

Anyways, I hope you see the importance of not relying entirely on your calculator! You should usually try doing some analysis of a problem without it.

19. Feb 5, 2005

### ktpr2

yeah im supposed to do math with my head, not my calculator. Live and learn I guess; the trig func should've tipped me off. Thanks all who contributed