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Given y^3+3y=x. Prove that y=f(x) is a single valued function in x.

Given y^3+3y=x. Prove that y=f(x) is a single valued function in x.

Grade:12

1 Answers

Aditya Gupta
2081 Points
3 years ago
Consider f(y)= y^3+3y-x
Let us assume x is a particular constant.
Then f'(y)= 3y^2+3 which is always greater than zero.
So since f(y) is a cubic in y, it is bound to have atleast one root, but since it is also an increasing function, hence there will be exactly one root.
Therefore for any real x there exists one and only real root of f(y). So f(y)=0 for single y only.
Or y^3+3y=x is a single valued function.

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