Friday

April 18, 2014

April 18, 2014

Posted by **Lynn** on Friday, March 28, 2008 at 5:20pm.

Problem

If f is one-to-one, find an equation for its inverse.

f(x)=x^3-5

y=x^3-5

x=y^3-5

5+x=y^3-5+5

5+x=y^3

My answer:

y^3=5+x

3y=35+x

y=35+x

f^-1(x)35+x

- Algebra Help Inverse One-to-one -
**Damon**, Friday, March 28, 2008 at 5:59pmy = (5+x)^(1/3)

is the easy way to write that.

Now lets say x = 3

then

y = 8^1/3 = 2

but what if y = (-1 + i sqrt 3)

square that

1 -2 i sqrt 3 - 3 = (-2 - 2i sqrt 3)

multiply that by (-1+ i sqrt 3)

2 -2 i sqrt 3 + 2 i sqrt 3 + 6

= 8 !!!!!

I also claim that the complex conjugate

y = (-1 -i sqrt 3)

cubed is also 8

therefore I claim that y = 2 or y = -1+i sqrt 3 or y = -1 -i sqrt 3

are all values of y for x = 3

So if your algebra subject is about complex as well as real numbers, there are three values of y for any value of x in the inverse problem.

- Algebra Help Inverse One-to-one -
**Lynn**, Friday, March 28, 2008 at 6:01pmSorry, my answer is:

y^3=5+x

3y=35+x

y=35+x

f^-1(x)=35+x

- Algebra Help Inverse One-to-one -
**Damon**, Friday, March 28, 2008 at 7:01pmbut whatever your symbol is for the cube root, you have to take the cube root of the whole right side, need parentheses

(5+x)

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