How do you find #f^-1(x)# given #f(x)=x^2-4x+3#?

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Answer 1 · 2016-12-11T05:26:18

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

We switch the x-y values by the definition of the inverse function (a reflection over the line #y =x#).

#x = y^2 - 4y + 3#

We now complete the square in order to solve for #y#.

#x = 1(y^2 - 4y + 4 - 4) + 3#

#x = 1(y - 2)^2 - 4 + 3#

#x= 1(y - 2)^2 - 1#

#x= (y - 2)^2 - 1#

Solve for #y#:

#x + 1 = (y - 2)^2#

#+-sqrt(x +1) = y - 2#

#2 +-sqrt(x + 1) = y#

Hopefully this helps!