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

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

thanks

Nov 2, 2017

#2
+7348
+1

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   =   271/3   =   3

Nov 2, 2017

#1
+556
+1

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)

Nov 2, 2017
#2
+7348
+1

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   =   271/3   =   3

hectictar Nov 2, 2017