Inverse Function

[Krash129]

Good Morning,
I am attempting to muddle my way through a question but I has been a while since I have dealt with something like:

The continuous function f(x + a) is even, where a > 0, and is defined for all real x.
Associated with the function f(x) is an inverse function f ^-1(x), which has a range (-∞,a].
If c > a, what is the value of f ^-1[f(x)] ?

A. 2a – c
B. a – c
C. – c
D. – a – c


I am told that the answer is A but cannot make the connection. I know that the symmetry of the even function should help when expressing c > a as a – c > 0
but then... ¯\_(ツ)_/¯

Many thanks in advance

Krash129

 

[Cubist]

Hi Krash129. I think there's a small typo in the question...

If c > a, what is the value of f ^-1[f( x)] ?

The variable I changed to red should probably be a "c" rather than an "x".

--

I personally find it helpful to add a new function g(x) where g(x) is even. (Note that f(x) is not even.) Since g(x) is even then g(x)=g(-x) , equation (1)

We are given f(x+a) is even, so let f(x+a)=g(x) , equation (2)

We are asked to find d=f ^-1[ f(c)] where c>a. equation (3)

(3) implies f(d) = f(c) where d<a due to the range of the inverse function

f(c) = g(c-a) ...obtained by putting c=x+a into (2)

= g( ??? ) ...using (1) can you write an expression for the "???"

= f( ??? ) ...use (2) again

= f(d) therefore d=???

 

[Steven G]

c > a as a – c > 0 ??

If a-c>0 shouldn't it be that a > c??

If you do not believe this, then add 'c' to both sides of a-c>0. What do you get?

 

[Krash129]

Hi Cubist,
Many thanks, especially for picking up the transcription error.
This was my thoughts after processing your setup...

f(c) = g(c-a)
= g(-c+a)
= f(-c+a+a)
=f(2a-c)
= f(d)

therefore d=2a-c

Let me know if I have mucked anything up. Otherwise many thanks.

 

[Krash129]

c > a as a – c > 0 ??

If a-c>0 shouldn't it be that a > c??

If you do not believe this, then add a to both sides of a-c>0. What do you get?

Yeah I see that now too, I think it may have been another transription error because it wasn't like that on my hand written work.
Cheers

 

[Cubist]

Hi Cubist,
Many thanks, especially for picking up the transcription error.
This was my thoughts after processing your setup...

f(c) = g(c-a)
= g(-c+a)
= f(-c+a+a)
=f(2a-c)
= f(d)

therefore d=2a-c

Let me know if I have mucked anything up. Otherwise many thanks.

Perfect! Well done.

The problem with this method, for me, is that I didn't get an "intuitive feel" that it's correct. But sometimes it's best to trust in the math. I guess that sketching a graph or plugging some numbers in might help to convince yourself if you feel the same way (perhaps simply choose g(x)=x^2).