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avatar+81 

Don't the domain and range of a function just swap for the inverse function?

Example-

F(x)-

Domain: [9, \(\infty\))

Range: (\(-\infty\),\(\infty\))

F-1(x)-

Domain: (\(-\infty\),\(\infty\))

Range: [9, \(\infty\))

Is that correct?

 Apr 4, 2019
edited by MemeLord  Apr 4, 2019
edited by MemeLord  Apr 4, 2019
 #3
avatar+104402 
+1

Not quite,

The range is often more limited for the inverse.

Further constrainst on range often have to be introduced because the inverse must also be a function

   Each x value must have AT MOST one y value.

 

eg

\(f(x)=sin x\)                  domain is all real x                  and range is      [-1,1]   

 

\(f^{-1}(x)=sin^{-1}x\)        domain  is    [-1,1]     but range must be limited. Uusally it is limited to    \([-\pi/2,\pi/2]\)

.
 Apr 4, 2019

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