Let $$f: R \longrightarrow R, f(x) = 4 - x$$

$f^{-1}( x ) =$

Let $$f: [0,\infty) \longrightarrow R, f(x) = 13 - x^2$$

$f^{-1}( x ) =$

Let $$f: C \longrightarrow D, f(x) = \frac { x + 2} { x + 9 }$$

f is bijective from C to D.

$f^{-1}( -6 ) =$

Let $$f(x) = \frac { 1 } { 4} x - 2 , \quad 2 \le x \le 9$$

The domain of $f^{-1}$ is the interval $[A,B]$

where $A =$ and where $B =$

Let $$f: R \longrightarrow R, f(x) = 4 + 2 x + 4 e^x$$ $f^{-1}(8) =$

If $f: R \longrightarrow R, f(x) = 5 x - 8$, then
$f^{-1}(y) =$
$f^{-1}( 11 ) =$