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avatar+2077 

\(f(n)=\sqrt[5]{n+2}\)

\(f(x)=-2x^3-1\)

Please help!

RainbowPanda  Oct 29, 2018
 #1
avatar+2077 
+3

I got the first one but need help with the second!

Let me know if you would like to see my work~

RainbowPanda  Oct 29, 2018
 #2
avatar+91027 
+3

Write the first as

 

y  =  ( n + 2)^(1/5)      take both sides to the 5th power

 

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

 

y^5  - 2  = n         "swap" n  and y

 

n^5  - 2  =  y      for y, write  f-1 (n)..so...

 

f-1(n)  = n^5  - 2

 

 

Second one 

 

y  = -2x^3  - 1      add 1 to both sides

 

y  + 1  =  -2x^3       multiply through by - 1

 

- (y + 1)  = 2x^3       divide both sides by 2

 

-(y + 1) / 2  =  x^3        take the cube root of both sides

 

[ - (y + 1)  / 2 ] ^(1/3)  =  x      "swap "  x  and y

 

[ - ( x + 1) / 2 ]^(1/3)  =  y      for y, write  f-1(x)

 

f-1(x)  =  [ - (x + 1) / 2 ] ^(1/3)

 

 

 

cool cool cool

CPhill  Oct 29, 2018
 #3
avatar+2077 
+2

Thank you

RainbowPanda  Oct 29, 2018
 #4
avatar+3270 
+2

Yes, instead of f(x), set it to y! 

tertre  Oct 29, 2018

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