Posted by Lynn on Friday, March 28, 2008 at 5:20pm.
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.
Sorry, my answer is:
y^3=5+x
3y=35+x
y=35+x
f^-1(x)=35+x
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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