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# Finding inverses

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

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

Oct 29, 2018

### 8+0 Answers

#1
+94215
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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

Oct 29, 2018
#2
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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
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Nope...doesn't matter.....a + b   = b +  a

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

CPhill  Oct 29, 2018
#3
+94215
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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)

Oct 29, 2018
#5
+2297
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I got $$g^-1(x)=\sqrt[5]{-x}$$

RainbowPanda  Oct 29, 2018
#6
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Same thing  ...I just wrote it in exponential form....good job  !!!

CPhill  Oct 29, 2018
#7
+2297
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Oh I see! My teacher does not want these in exponential form >.<

RainbowPanda  Oct 29, 2018
#8
+94215
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OK.....I'll put them in  radical  form

CPhill  Oct 29, 2018

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