+0  
 
0
41
8
avatar+2051 

\(f(x)=\sqrt[3]\frac{x-2}{2}\)

\(g(x)=-x^5\)

RainbowPanda  Oct 29, 2018
 #1
avatar+90968 
+2

The idea is to get x by itself....then "swap" x and y...and for y, write the inverse 

 

y  =  [  ( x - 2 ) / 2 ]^(1/3)     cube both sides

 

y^3  = ( x - 2) / 2      multiply both sides by 2

 

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

 

2y^3 + 2  = x      "swap"  x and  y

 

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

 

f-1(x)  = 2x^3 + 2

 

cool cool cool

CPhill  Oct 29, 2018
 #2
avatar+2051 
+1

That's what I got for #1 but I wrote it like 2+2x^3 dont think that matters though

RainbowPanda  Oct 29, 2018
 #4
avatar+90968 
+1

Nope...doesn't matter.....a + b   = b +  a

 

[addition is  commutative...we can write  it in any order  ]

 

 

cool cool cool

CPhill  Oct 29, 2018
 #3
avatar+90968 
+2

y  = -x^5      multiply through by -1

 

-y = x^5      take the 5th  root of both sides

 

[-y ]^(1/5)  =  x      "swap" x and y

 

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

 

g-1(x)  = (-x)^(1/5)

 

 

cool cool cool

CPhill  Oct 29, 2018
 #5
avatar+2051 
+1

I got \(g^-1(x)=\sqrt[5]{-x}\)

RainbowPanda  Oct 29, 2018
 #6
avatar+90968 
+1

Same thing  ...I just wrote it in exponential form....good job  !!!

 

 

cool cool cool

CPhill  Oct 29, 2018
 #7
avatar+2051 
+1

Oh I see! My teacher does not want these in exponential form >.<

RainbowPanda  Oct 29, 2018
 #8
avatar+90968 
+1

OK.....I'll put them in  radical  form

 

 

cool cool cool

CPhill  Oct 29, 2018

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