Inverse of a function

  • Thread starter hangover
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  • #1
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If a function is bijective, then its inverse exists. Is there any example that inverse of a function exists but the original function is not bijective?
 

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  • #2
radou
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A function has to be bijective in order to have an inverse.
 
  • #3
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A function can be locally bijective, so it's inverse exists only in some finite interval.

For example [tex]x^{2}[/tex] is not a bijective in any interval containing x=0 (since f'(0)=0) but if you restrict yourself to x>0, then you off course have the inverse
[tex]f(x)=\sqrt{x}[/tex] or in x<0 the inverse is [tex]f(x)=-\sqrt{x}[/tex].
 
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  • #4
HallsofIvy
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Another example. f(x)= ex, as a function from R to R, is not surjective, so not and does not have an inverse. In particular, there is no f-1(-1). But if I consider it as a function from R to R+, the positive real numbers, then it is bijective and f-1(x)= ln(x).

A function, from A to B, has an inverse if and only if it is bijective.
 
  • #5
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Thanks a lot! It may be a typo in my textbook.

A function can be locally bijective, so it's inverse exists only in some finite interval.

For example [tex]x^{2}[/tex] is not a bijective in any interval containing x=0 (since f'(0)=0) but if you restrict yourself to x>0, then you off course have the inverse
[tex]f(x)=\sqrt{x}[/tex] or in x<0 the inverse is [tex]f(x)=-\sqrt{-x}[/tex].
However, why is x^2 not bijective if we define the domain containing zero(like x=>0)? It is a continous function so isn't it bijective at this interval though being not strictly increasing? Thanks
 
  • #6
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f(x) = x2 on [0, infinity) is strictly increasing. If a and b are any two numbers in this interval such that a < b, then f(a) < f(b).
 
  • #7
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Thanks a lot! It may be a typo in my textbook.



However, why is x^2 not bijective if we define the domain containing zero(like x=>0)? It is a continous function so isn't it bijective at this interval though being not strictly increasing? Thanks

I had a little confusion in defining the inverse sorry.

For [tex]x\geq 0[/tex] the inverse of [tex]y=x^{2}[/tex] is [tex]x=f^{-1}(y)=\sqrt{y}[/tex]


For [tex]x\leq 0[/tex] the inverse of [tex]y=x^{2}[/tex] is [tex]x=f^{-1}(y)=-\sqrt{y}[/tex]

Notice of course that the inverse is defined only over [tex]y\geq 0[/tex], since the range of x^2 is only the non-negative real numbers.

Also, the interval where the function can be define as a bijection may inclue x=0, but only as a boundary point.
 
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