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avatar+2442 

Last ones that I need help with for today :)

\(f(x)=\sqrt[3]{x-1}\)

\(g(n)=2+2n^5\)

 Oct 29, 2018
 #1
avatar+98044 
+2

y  =  ( x - 1)^(1/3)    cube both sides

y^3  = x  - 1      add 1 to both sides

y^3 + 1  =  x    "swap"  x an y

x^3 + 1  =  y      for y, write  f-1(x)

f-1(x)  = x^3 + 1

 

y  = 2 + 2n^5    subtract 2 from both sides

y - 2  =  2n^5     divide both sides by 2

[ y - 2 ] / 2  = n^5     take the 5th  root

( [ y - 2 ] / 2 ) ^(1/5)  = n       "swap"  n and y

( [ n - 2 ]  / 2 ) ^(1/5)  =  y     for y, write g-1(n)

g-1(n)  =  ( [ n - 2 ]  / 2 ) ^ (1/5)   =   5√ [  ( n - 2) / 2 ]

 

 

cool cool cool

 Oct 29, 2018

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