If f(x)=x^3-8, what is f^{-1}(f(f^{-1}(19)))?

f( f^-1(x)  )   =   x

Therefore

f( f^-1(19) )   =   19

And

f^-1(   f( f^-1(19) )   )   =   f^-1( 19 )

To find  f^-1(19)  , lets first find  f^-1(x) .

f(x)  =  x³ - 8

y  =  x³ - 8     →     y + 8  =  x³     →     (y + 8)^1/3   =  x

f^-1(x)  =  (x + 8)^1/3      Now plug in  19  into this function.

f^-1(19)  =  (19 + 8)^1/3   =   27^1/3   =   3

I'm assuming f^{-1} is a way of switching the input and output values, which works for this function. Therefore f^{-1}(f(f^{-1(19))) --> f^{-1} = 19, so f(x) = 19. I substituted the equation in and got x^3-8 = 19. I added 8 to each side to get x^3 = 27. To get x without an exponent, I took the cube root of each side, getting x = 3. Either way, f^{-1}(f(f^{-1(19))) = f^{-1}(19) = f(3)