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?

MemeLord Apr 4, 2019

#3**+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 \([-\pi/2,\pi/2]\)

.Melody Apr 4, 2019