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

  • Thread starter ktpr2
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In summary, the teacher says f(f_inverse(5)) = 5, but I think it's undefined. I don't think the teacher considered the domain of f_inverse, so who's wrong? The domain and range of f^-1 is the range of f, which doesn't include 5. If f has an inverse and a is in the range of f... then f(f_inverse(a))=a. But 5 is not a member of the range of f_inverse, so f(f_inverse(5)) must be undefined.
  • #1
ktpr2
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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?
 
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  • #2
What's the domain of f^-1?
 
  • #3
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
ktpr2 said:
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.
 
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  • #5
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
ktpr2 said:
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.
 
  • #7
What's the range of tan (pi x/2) for -1 < x < 1?
 
  • #8
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
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
Hurkyl said:
What's the range of tan (pi x/2) for -1 < x < 1?

(Without using your calculator)
 
  • #11
Take the calculator & compute the alue for x=0.999.

Daniel.
 
  • #12
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
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
Okay,Hurkyl,i'll leave it up to you... :wink:

Daniel.
 
  • #15
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.
 
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  • #16
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)?
 
  • #17
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
 
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  • #18
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
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
 

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

1. What does it mean when F (f inverse (5)) = undefined?

When F (f inverse (5)) is undefined, it means that the inverse function of f does not exist at the input value of 5. In other words, there is no real number x that, when plugged into f, would result in an output of 5. This could be due to various reasons such as the function being undefined at that point or the function not being one-to-one.

2. How can I determine if F (f inverse (5)) is undefined?

To determine if F (f inverse (5)) is undefined, you can start by finding the inverse function of f (f inverse) and then plugging in the input value of 5. If the result is undefined or does not exist, then F (f inverse (5)) is also undefined. Additionally, you can also check for any restrictions or conditions given in the original function f that would make the inverse function undefined at that point.

3. Can a function have an undefined inverse at more than one input value?

Yes, a function can have an undefined inverse at more than one input value. This can happen if the function has multiple points of discontinuity or if there are restrictions or conditions that make the inverse function undefined at multiple points. It is important to carefully analyze the original function and its inverse to determine all the possible points of undefined inverse.

4. What is the difference between an undefined inverse and an inverse that does not exist?

An undefined inverse and an inverse that does not exist are essentially the same thing. Both indicate that there is no real number that can be plugged into the function to result in the given output. The only difference may be in the terminology used, where "undefined inverse" may imply that there is a possibility for the inverse to exist but is undefined at that particular input value, while "inverse does not exist" may imply that there is no inverse function for the given function at all.

5. Can a function have an inverse if it has an undefined inverse at any point?

No, a function cannot have an inverse if it has an undefined inverse at any point. This is because for a function to have an inverse, it must be one-to-one and have a unique output for every input. If there are any points where the inverse is undefined, it means that the function is not one-to-one and therefore does not have a true inverse function. However, a function can have an inverse if it has discontinuities or restrictions that make the inverse undefined at certain points, as long as it is still one-to-one overall.

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