Who's Correct About f(f_inverse(5))?

[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?

 

[Hurkyl]

What's the domain of f^-1?

 

[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?

 

[learningphysics]

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?

Your teacher is right.

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.

 

[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?

 

[learningphysics]

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?

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

 

[Hurkyl]

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

 

[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.

 

[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.

 

[Hurkyl]

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

(Without using your calculator)

 

[dextercioby]

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

Daniel.

 

[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.

 

[Hurkyl]

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?

 

[dextercioby]

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

Daniel.

 

[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 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.

 

[Hurkyl]

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

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)?

 

[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

 

[Hurkyl]

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.

 

[ktpr2]

yeah I am 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