Posted by Lynn on Friday, March 28, 2008 at 5:20pm.
I need to know if I have done this correctly.
Problem
If f is onetoone, find an equation for its inverse.
f(x)=x^35
y=x^35
x=y^35
5+x=y^35+5
5+x=y^3
My answer:
y^3=5+x
3ãy=3ã5+x
y=3ã5+x
f^1(x)3ã5+x

Algebra Help Inverse Onetoone  Damon, Friday, March 28, 2008 at 5:59pm
y = (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 Onetoone  Lynn, Friday, March 28, 2008 at 6:01pm
Sorry, my answer is:
y^3=5+x
3ãy=3ã5+x
y=3ã5+x
f^1(x)=3ã5+x

Algebra Help Inverse Onetoone  Damon, Friday, March 28, 2008 at 7:01pm
but 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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