Make "x" the subject (1 Viewer)

[lyounamu]

The answer in the book is:

g^-1(x) = 1-sqrt(1-x²)/x (i assume they just divided everything by 2)

I wonder why they only took the negative root and left out the other consideration?

Our answers are same as your answer. In the answer, they just took out root 4 out which is basically 2. And divide everything by 2.

EDIT: x + xy^2 = 2y

xy^2 - 2y +x =0

a=x, b= -2 and c=x

You use the quadratic formula

Therefore, y= (2 +_ root of (4-4x^2))/2x
= ((2+_ root of (4) . root of (1-x^2))/2x
= (2+ _ 2root of (1-x^2))/2x
= (1 +_ root of (1-x^2))/x

 

[lacklustre]

Our answers are same as your answer. In the answer, they just took out root 4 out which is basically 2. And divide everything by 2.

EDIT: x + xy^2 = 2y

xy^2 - 2y +x =0

a=x, b= -2 and c=x

You use the quadratic formula

Therefore, y= (2 +_ root of (4-4x^2))/2x
= ((2+_ root of (4) . root of (1-x^2))/2x
= (2+ _ 2root of (1-x^2))/2x
= (1 +_ root of (1-x^2))/x

Cheers man.


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[3unitz]

The answer in the book is:

g^-1(x) = 1-sqrt(1-x²)/x (i assume they just divided everything by 2)

I wonder why they only took the negative root and left out the other consideration?

inverse functions 1 to 1