What is the inverse of f(x)=x^2-4x+1 ?

The inverse of f(x)=x^2-4x+1 is 2+sqrt(x+3),2-sqrt(x+3)

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inverse of f(x)=x^2-4x+1

inverse

f (x) = x^{2} - 4 x + 1

2 + x + 3, 2 - x + 3

Solution steps

y = x^{2} - 4 x + 1

Replace

x

with

y

x = y^{2} - 4 y + 1

Solve for

y, x = y^{2} - 4 y + 1

2 + x + 3, 2 - x + 3

f (x) = \frac{x + 18}{x - 16}

\frac{x ^{2} + 3 x - 10}{x - 3}

\frac{x - 1}{x ^{3} + 4 x ^{2} + 12 x + 8}

f (x) = \frac{sec ( π x )}{2}

2 x^{6} + x^{8} + 3

What is the inverse of f(x)=x^2-4x+1 ?
The inverse of f(x)=x^2-4x+1 is 2+sqrt(x+3),2-sqrt(x+3)