What is the inverse function of #f(x)=x^2#?

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Answer 1 · 2015-09-11T00:11:46

The inverse is #f^-1(x)=sqrtx#

Hence #f(x)=x^2=>y=x^2=>sqrty=sqrt(x^2)=>sqrty=absx=>x=+-sqrty#

Answer 2 · 2015-09-11T01:33:27

#f(x) = x^2# is not one-to-one. It does not have an inverse function.

If we try to solve #y=x^2# for #x# we do not get a single value. That means we do not get a function.

We get #x = +- sqrty#.

In order to be invertible a function must be one-to-one.

That means that we must have:
for every #x_1 != x_2#, we get #f(x_1) != f(x_2).#.

For #f(x) = x^2#, we ghave #f(-1) = f(1)# (for example), so there is no inverse function.