How do you find #f^-1(x)# given #f(x)=(x-3)^2+7#?

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Answer 1 · 2017-01-31T11:20:40

The answer is #f^-1(x)=3+sqrt(x-7)#

Let #y=(x-3)^2+7#

#(x-3)^2=y-7#

#x-3=sqrt(y-7)#

#x=sqrt(y-7)+3#

Therefore,

#f^-1(x)=3+sqrt(x-7)#

Verification,

#f(f^-1(x))=f(3+sqrt(x-7))=(3+sqrt(x-7)-3)^2+7#

#=(sqrt(x-7))^2-7#=x-7+7=x