What is the inverse of #f(x) = 1 - x^2, x>= 0#?

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Answer 1 · 2017-09-19T14:33:52

The inverse is #=sqrt(1-x)#

Our function is #f(x)=1-x^2# and #x>=0#

Let

#y=1-x^2#

#x^2=1-y#

Exchanging the #x# and #y#

#y^2=1-x#

#y=sqrt(1-x)#

Therefore,

#f^-1(x)=sqrt(1-x)#

Verification

#[fof^-1] (x)=f(f^-1(x))=f(sqrt(1-x))=1-(sqrt(1-x))^2=1-1+x=x#

graph{(y-1+x^2)(y-sqrt(1-x))(y-x)=0 [-0.097, 2.304, -0.111, 1.089]}